Supplemental Guide: Calculating pH at Different Points in a Titration

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Solving for pH at different points in an acid-base titration can be quite tricky because the method changes somewhat depending on the type of titration and sometimes where you are in the titration. A good general method is as follows:

Step 1: Write and balance the reaction equation for the titrant (usually a strong base) reacting with the analyte (usually a weak or strong acid)

Example: $LaTeX: HA+NaOH\rightarrow NaA+H_2O$ $HA+NaOH\rightarrow NaA+H_2O$

Step 2: Calculate the initial amount of moles or millimoles of each reactant.

Step 3: Set up an ICF chart (Initial, Change, Final). NOTE: An ICF chart uses moles or millimoles, NOT molarity.

Step 4: Use the ICF chart to determine the final amounts of each reactant and product. (Remember that a strong species will drive a reaction to completion, so look for which reactant is your limiting reagent.)

Titration img 2.png

Step 5: Depends on type of titration.

For a Strong Acid-Strong Base titration, there are three possibilities:

1. 1. 1. If there is excess HA at the end of the reaction, calculate its new molarity. Then, use $LaTeX: pH=-\log\left[H_3O^+\right]$ $pH=-\log\left[H_3O^+\right]$ to solve for pH.
    2. If there is excess OH^- at the end of the reaction, calculate its new molarity. Then, calculate the pOH $LaTeX: \left(pOH=-\log\left[OH^-\right]\right)$ $\left(pOH=-\log\left[OH^-\right]\right)$ . Then, calculate the pH. (Remember that pH + pOH = 14)
    3. If there is no HA and OH^- at the end of the reaction, the solution has reached the equivalence point and should have a pH of about 7.

Additional Considerations for Weak Acid-Strong Base Titrations

For a Weak Acid-Strong Base titration, solving for the pH requires a different process depending on what region you are in. The four regions labeled in the following graph and methods for calculating pH in each one are described below. Use the ICF chart in STEP 4 to determine which region applies to the given situation:

Initial Region: Skip steps 1-4. At this point, no strong base has been added. To determine the pH, write the equilibrium reaction equation of the weak acid (HA) reacting with water, set up an ICE chart, and use the K_a to solve for the [H₃O⁺]. Then, use $LaTeX: pH=-\log\left[H_3O^+\right]$ $pH=-\log\left[H_3O^+\right]$ .

Buffer Region: At this point, there is both weak acid (HA) and weak conjugate base (A^-) present. Therefore, the Henderson-Hasselbalch equation $LaTeX: \left(pH=pK_a+\log\frac{\left[base\right]}{\left[acid\right]}\right)$ $\left(pH=pK_a+\log\frac{\left[base\right]}{\left[acid\right]}\right)$ along with the K_a of the weak acid can be used to solve for pH.

Equivalence point: At this point, all of the weak acid has been neutralized by an equal amount of strong base. As such, the only pH affecting species present is weak conjugate base (A^-). To solve for pH, first write the equilibrium reaction equation of the weak base (A^-) reacting with water and calculate K_b for the weak base. Then, set up an ICE chart and calculate [OH^-] using the K_b. Then, calculate the pOH $LaTeX: \left(pOH=-\log\left[OH^-\right]\right)$ $\left(pOH=-\log\left[OH^-\right]\right)$ . Then, calculate the pH. (Remember that pH + pOH = 14)

Excess Region: At this point, there is both weak base (A^-) and excess strong base (OH^-) present in the solution. Since the effect of the strong base is going to be so much stronger than the weak base, the weak base can be ignored when calculating pH. To solve for pH, determine the amount of excess moles of strong base, divide by the total volume (in Liters) to get [OH^-], and then calculate the pOH. Then, calculate pH.