17.3 pH Calculations Involving Titrations
PRE-HEALTH PREP
Strong Acid-Strong Base Titrations
pH Calculations at Every Point on the Titration Curve
General Chemistry 17.3
Lesson Overview
Strong acid-strong base titrations involve three distinct pH calculation types corresponding to regions before, at, and beyond the equivalence point. Each calculation treats the neutralization reaction as a limiting reagent problem, converting volumes and molarities to moles to determine which reactant remains in excess. Worked examples using HCl and NaOH demonstrate the complete process, including the additional pOH-to-pH conversion step required when excess strong base is present.
The Titration Curve and 3 pH Calculation Types
In a strong acid-strong base titration, there are really only 3 types of pH calculations you will ever encounter. Each corresponds to a different region of the titration curve: before the equivalence point, at the equivalence point, and after the equivalence point. Once you understand the logic behind each one, you can calculate pH at any point on any strong acid-strong base titration curve.
Consider the setup for this lesson: we begin with 150 mL of 0.1 M HCl and progressively add 0.125 M NaOH. We will calculate the pH at 4 key points along the titration curve — the initial point, one point before equivalence, the equivalence point itself, and one point beyond it — giving you a complete picture of every possible calculation type.
| Region | Solution Character | pH Calculation Strategy |
|---|---|---|
| Before equivalence point | Acidic (excess HCl) | pH = −log[H+] |
| At equivalence point | Neutral | pH = 7 (nothing left to affect pH) |
| Beyond equivalence point | Basic (excess NaOH) | pOH = −log[OH−], then pH = 14 − pOH |
Strong Acid-Strong Base Titrations Only
The equivalence point pH equals exactly 7 only for strong acid-strong base titrations. In weak acid-strong base or weak base-strong acid titrations, the equivalence point pH shifts away from 7. That distinction matters on exams.
Acid-Base Neutralization as a Limiting Reagent Problem
When a strong acid reacts with a strong base, the neutralization reaction goes to completion. There is no equilibrium to worry about — one reactant runs out entirely. This makes every strong acid-strong base titration calculation a classic limiting reagent problem, and the key quantity you need is moles, not concentration.
For reactions at equilibrium, you work with molarities because equilibrium constants are expressed as ratios of concentrations. But for reactions that go to completion, identifying which reagent runs out first is all that matters — and that requires knowing how many moles of each you have.
For all our titration calculations we will follow the same general procedure:
- Calculate the number of moles of acid and base.
- Have the neutralization reaction go to completion.
- Based upon what is left after the neutralization reaction, calculate the pH.
There are 5 types of solutions you are required to be able to calculate the pH of (strong acid, strong base, weak acid, weak base, and buffer). These were presented in earlier chapters and are briefly summarized in the following reference image:
After neutralization is complete in any titration the pH of the resulting solution will be calculated using one of these 5 methods.
Converting to Moles
Recall that moles = molarity × volume (in liters). This rearrangement of the molarity equation (M = mol / L) is the starting point for every pH calculation in a strong acid-strong base titration.
Initial Point: 0 mL NaOH Added (pH = 1)
Before any base is added, the flask contains only 0.1 M HCl. Since HCl is a strong acid, it dissociates completely in water. The H+ concentration equals the initial HCl concentration, and the pH calculation is immediate.
pH = −log(0.1) = 1
This gives us the starting point of the titration curve at pH = 1.
Where the Titration Curve Starts Depends on Concentration
The initial pH of a strong acid-strong base titration curve can start anywhere below 7, depending on the acid concentration. A higher concentration means a lower starting pH. If you were titrating a base with an acid instead, the curve would start above 7.
Asymptotic Behavior at the End of the Curve
At the far end of the titration curve, pH approaches the value you would calculate for a pure solution of the titrant (in this case, 0.125 M NaOH). However, it never quite reaches that value because some of the hydroxide is always consumed by neutralization with HCl.
Before the Equivalence Point: 100 mL NaOH Added (pH = 2)
After adding 100 mL of 0.125 M NaOH to the original 150 mL of 0.1 M HCl, we need to determine how much acid remains after the neutralization reaction runs to completion. This is the standard 3-step process used for every pre-equivalence pH calculation.
Step 1: Calculate Moles of Each Reactant
Moles of NaOH = 0.125 M × 0.100 L = 0.0125 mol
Step 2: Identify the Limiting Reagent
Since we have more moles of HCl (0.015) than NaOH (0.0125), NaOH is the limiting reagent and is completely consumed. HCl and NaOH react in a 1:1 molar ratio, so both lose 0.0125 mol.
What About the Products?
The reaction also produces water and NaCl, but neither affects the pH. The solution is already aqueous, so a tiny bit more water is negligible. And NaCl is a neutral salt — Na+ is a Group 1 cation (negligible acidity) and Cl− is the conjugate base of a strong acid (negligible basicity). We can safely ignore both.
Step 3: Convert Moles to Concentration and Calculate pH
After the reaction, the only species affecting pH is the excess HCl. To find its concentration, divide the remaining moles by the total volume of the mixed solution.
[H+] = 0.0025 mol ÷ 0.25 L = 0.01 M
pH = −log(0.01) = 2
Why Use Total Volume?
When two solutions are mixed, the excess reagent is now dissolved in the combined volume of both solutions. Forgetting to add the volumes together is a common mistake on titration pH calculation problems. Always use the total volume when converting moles back to molarity.
Weak Acid-Strong Base Titrations Are Different
In a strong acid-strong base titration, the salt produced (e.g., NaCl) is neutral and does not affect pH. In weak acid-strong base or weak base-strong acid titrations, the salt will be acidic or basic and must be accounted for. That added complexity is covered in the weak acid/strong base and weak base/strong acid sections below.
At the Equivalence Point: 120 mL NaOH Added (pH = 7)
When 120 mL of 0.125 M NaOH has been added, the moles of base exactly equal the moles of acid. This is the equivalence point of the strong acid-strong base titration.
Moles of NaOH = 0.125 M × 0.120 L = 0.015 mol
Both reactants are completely consumed. After the neutralization reaction finishes, nothing remains in solution that can affect the pH. The NaCl produced is a neutral salt with no influence on pH. With no acid or base left, the solution is simply neutral.
No pH Calculation Needed at the Equivalence Point
At the equivalence point of a strong acid-strong base titration, there is no pH calculation to perform. Both the acid and base have completely neutralized each other, leaving only water and a neutral salt. The equivalence point pH is simply 7.
This is the inflection point on the titration curve — the steepest part of the S-shaped curve where a tiny addition of base causes a dramatic pH change.
Beyond the Equivalence Point: 150 mL NaOH Added (pH = 12.1)
For the final data point, we add 150 mL of NaOH. You might notice that this is the same volume as the original HCl solution. However, the NaOH concentration (0.125 M) is higher than the HCl concentration (0.1 M), so there will be more moles of base than acid — placing us firmly beyond the equivalence point. Post-equivalence pH calculations require an extra step compared to pre-equivalence calculations.
Step 1: Calculate Moles
Moles of NaOH = 0.125 M × 0.150 L = 0.01875 mol
Step 2: Find the Excess Reagent
This time HCl is the limiting reagent and is completely consumed. In the 1:1 reaction, both reactants lose 0.015 mol.
Step 3: Calculate pOH, Then Convert to pH
With excess strong base remaining, we first calculate the hydroxide ion concentration, then find pOH, and finally convert to pH using the relationship pH + pOH = 14.
[OH−] = 0.00375 mol ÷ 0.30 L = 0.0125 M
pOH = −log(0.0125) = 1.9
pH = 14 − 1.9 = 12.1
The Extra Step Beyond the Equivalence Point
Before the equivalence point, you go directly from [H+] to pH. Beyond the equivalence point, there is an extra step: calculate pOH first from [OH−], then subtract from 14 to get pH. This two-step pOH-to-pH conversion is the key difference between pre- and post-equivalence titration calculations.
Titration of a Strong Base with a Strong Acid
For the titration of a strong base with a strong acid the titration curve would be inverted but the pH calculations would be very similar to the titration of a strong acid with a strong base.
pH Calculations for Weak Acid / Strong Base Titrations
The pH calculations for the titration of weak acid with a strong base are generally more difficult than those for a strong acid with a strong base. One reason is that the salt formed in the neutralization reaction can't be ignored as it will be a basic salt rather than a neutral salt and will affect the pH of the solution. There are 4 areas on the curve requiring different methods of calculating the pH:
- 1. Initial Point
- 2. Before the Equivalence Point (excluding initial point)
- 3. At the Equivalence Point
- 4. After the Equivalence Point
Here, the base is strong. Because of this the resulting neutralization reaction will go to completion, and because it goes to completion we will also treat this as a limiting reagent problem. We will follow the same general procedure outlined above: (1) calculate moles, (2) run the neutralization to completion, (3) calculate pH from what remains.
For these examples specifically consider the titration of 100 mL of 0.10 M HF with 0.10 M NaOH. For these calculations Ka for HF = 6.8×10−4.
1. Initial Point
Before the addition of any NaOH we just have a solution of a weak acid. In Lesson 16.4 pH Calculations for Weak Acids and Bases, we learned a shortcut for calculating the pH of a weak acid solution:
It turns out that this corresponds to a percent ionization of ~8%. Being more than 5% this probably should have been solved using the quadratic equation rather than the shortcut. But even had we used the quadratic the calculated pH still would have been 2.1 (2.10 vs 2.06) so the error here is minimal.
2. Before the Equivalence Point (excluding initial point)
Addition of 25 mL 0.10 M NaOH
(0.10M)(0.025L) = 0.0025 moles NaOH
With the moles of acid and base determined, we run the neutralization reaction between HF and NaOH to completion.
HF and NaOH react in a 1:1 ratio, and with fewer moles of NaOH it is the limiting reagent. After the neutralization reaction has run to completion we can see that there will be 0.0075 moles of excess HF remaining and 0.0025 moles of NaF formed. We now have a weak acid and its conjugate base present in solution and have a buffer solution. Technically, we could use the Ka or Kb expressions to solve for the pH, but it's much more common to use the Henderson-Hasselbalch equation instead. (Even if outside of the buffer region which requires within a 1:10 or 10:1 ratio of weak acid / conjugate base we could still use the Henderson-Hasselbalch equation to calculate the pH before the equivalence point.)
The Henderson-Hasselbalch equation can be expressed using either molarities or moles (n) in the ratio of A− to HA—they are equivalent. While the equation is normally presented using molarities, it is much more convenient here to use moles so that we don't have to convert the moles of HF and NaF back into molarities.
2. Half-Equivalence Point (still Before the Equivalence Point)
One special point that falls in the range of 'before the equivalence point' is the half-equivalence point. At this point [A−] = [HA] and pH = pKa. If you can recognize you're at this point early on you can save a lot of time, but if you work it out you'll still arrive at the same answer, pH = pKa.
Addition of 50 mL 0.10 M NaOH
(0.10M)(0.05L) = 0.005 moles NaOH
Again, we run the neutralization to completion to see what remains.
NaOH is still the limiting reagent. After the neutralization reaction has run to completion we can see that there will be 0.005 moles of excess HF remaining and 0.005 moles of NaF formed. We still have a buffer solution and will still use the Henderson-Hasselbalch equation to calculate the pH. With the moles of F− and HF being equal, the ratio of moles F− to moles HF equals 1 and log(1) = 0, so pH = pKa.
3. At the Equivalence Point
Addition of 100 mL 0.10 M NaOH
(0.10M)(0.10L) = 0.01 moles NaOH
Notice that the moles of HF and NaOH are equal — this is the equivalence point. The neutralization reaction goes to completion with neither reactant left over.
At this point the HF and NaOH have completely neutralized each other and the only thing remaining that will affect the pH is NaF, a weak base. In Lesson 16.4 pH Calculations for Weak Acids and Bases, we learned a shortcut for calculating the pH of a weak base solution:
But to use this equation we'll first have to calculate the Kb of F− from the Ka of HF. Since Ka × Kb = Kw, then Kb = Kw / Ka. This equation also requires us to plug in the molarity of the weak base. At this point we only know that we have 0.01 moles of weak base but we'll have to divide that by the total liters of solution to convert it into a molarity. As 100 mL HF and 100 mL NaOH have been combined, the total volume is 200 mL, or 0.2 L.
Why pH > 7 at the Equivalence Point?
Unlike a strong acid-strong base titration where the equivalence point is pH = 7, the equivalence point of a weak acid-strong base titration is above 7. That is because the only species remaining at the equivalence point is a weak base (the conjugate base of the weak acid), which causes the solution to be basic.
4. After the Equivalence Point
Addition of 120 mL 0.10 M NaOH
(0.10M)(0.12L) = 0.012 moles NaOH
This time, NaOH is in excess. The neutralization still goes to completion, but now HF is the limiting reagent.
After the equivalence point we have excess NaOH, a strong base, remaining in the solution. We have also formed NaF, a weak base. As a strong base dissociates completely, whereas a weak base dissociates to a small extent (usually <5%), we generally ignore the weak base and calculate the pOH as if it were a solution of a strong base alone. In this example we have 0.002 moles of NaOH now in a total volume of 220 mL, or 0.22 L.
pH Calculations for Weak Base / Strong Acid Titrations
The pH calculations for the titration of weak base with a strong acid are very similar to those of a weak acid with a strong base. While not worked out explicitly here, the following curve shows the four areas involving a different way to calculate the pH and how to do so:
- 1. Initial Point
- 2. Before the Equivalence Point (excluding initial point)
- 3. At the Equivalence Point
- 4. After the Equivalence Point
Key Difference: Equivalence Point pH < 7
For a weak base / strong acid titration the equivalence point occurs at a pH below 7, because the only species remaining is the conjugate acid of the weak base, which is acidic. The same limiting reagent approach applies throughout, but the Henderson-Hasselbalch equation is used before the equivalence point and a weak acid calculation is used at the equivalence point.
Summary: All 3 Titration pH Calculation Types
Every pH calculation for a strong acid-strong base titration falls into one of 3 categories. Here is the complete decision framework:
| Point on the Curve | What's Left in Solution? | How to Find pH |
|---|---|---|
| Before equivalence | Excess strong acid (HCl) | Find excess moles of H+ → divide by total volume → pH = −log[H+] |
| At equivalence | Neutral salt only (NaCl) | pH = 7 (no calculation needed) |
| After equivalence | Excess strong base (NaOH) | Find excess moles of OH− → divide by total volume → pOH = −log[OH−] → pH = 14 − pOH |
The Unifying Idea for All Titration pH Calculations
Regardless of where you are on the titration curve, the method is the same: treat the neutralization as a limiting reagent problem, determine what is left over, then use the appropriate pH formula. The only thing that changes is whether you are working with excess acid (use pH directly) or excess base (find pOH first, then convert using pH = 14 − pOH).
Strong vs. Weak Titrations: Key Difference
The strong acid-strong base examples above all share a simplifying feature: the salt produced (NaCl) is neutral and does not affect pH. The weak acid/strong base and weak base/strong acid sections that follow show what changes when the salt is acidic or basic — including buffer regions, shifted equivalence points, and equilibrium calculations.
9 Key Takeaways
- Strong acid-strong base titrations involve only 3 types of pH calculations: before, at, and after the equivalence point.
- Treat every neutralization as a limiting reagent problem. Convert volumes and molarities to moles, determine which reagent runs out first, and find what remains.
- Before the equivalence point, excess strong acid remains. Calculate [H+] from the excess moles divided by total volume, then take pH = −log[H+].
- At the equivalence point, moles of acid and base are equal. Only neutral salt (NaCl) remains, so pH = 7 with no calculation required.
- After the equivalence point, excess strong base remains. Calculate [OH−] from excess moles divided by total volume, find pOH = −log[OH−], then convert with pH = 14 − pOH.
- Always use total volume (acid + base volumes combined) when converting excess moles back to molarity.
- NaCl is a neutral salt and does not affect pH. This simplification applies only to strong acid-strong base titrations — weak acid or weak base titrations produce salts that do affect pH.
- Weak acid/strong base equivalence point pH > 7 because the conjugate base of the weak acid is itself a weak base, making the solution basic at the equivalence point.
- Weak base/strong acid equivalence point pH < 7 because the conjugate acid of the weak base is itself a weak acid, making the solution acidic at the equivalence point.