Question

A solution is made by adding 70.00grams of sodium acetate to 400.0mL of 0.60 mol/L solution of acetic acid.
a.What is the pH of the solution?
b.10.0mL of 2.0M of sodium hydroxide solution was added to the solution above. Calculate the new pH.
c.10.0mL of 2.0M of hydrochloric solution was added to the solution in part a. Calculate the new pH.

Answer 1

Question #67345, Chemistry

A solution is made by adding 70.00 grams of sodium acetate to 400.0 mL of 0.60 mol/L solution of acetic acid.

a. What is the pH of the solution?

System described above is the buffer solution. pH of such kind of solutions calculates by the following formula:

$$pH = 7 + \frac{1}{2} pK_a + \frac{1}{2} \lg C_{NaAc}$$

where

$$\begin{array}{l} pK_a = -\lg K_{HAc} = 4.76 \text{ [1]} \\ C_{NaAc} = \frac{m(NaAc)}{M(NaAc) * V} = \frac{70}{82.03 * 0.4} = 2.13 \text{ mol/L} \\ pH = 7 + \frac{1}{2} * 4.76 + \frac{1}{2} \lg 2.13 = 9.5441 \\ \end{array}$$

b. 10.0 mL of 2.0 M of sodium hydroxide solution was added to the solution above. Calculate the new pH.

So, concentration of sodium acetate was changed:

$$\begin{array}{l} C'_{NaAc} = \frac{C_{NaAc} * V + C_{NaOH} * V_{NaOH}}{V + V_{NaOH}} = \frac{2.13 * 0.4 + 0.01 * 2}{0.41} = 2.126 \\ pH = 7 + \frac{1}{2} * 4.76 + \frac{1}{2} \lg 2.126 = 9.5438 \\ \end{array}$$

c. 10.0 mL of 2.0 M of hydrochloric solution was added to the solution in part a. Calculate the new pH.

In this case pH calculates with the following formula:

$$\begin{array}{l} pH = 7 + \lg C'_{HCl} + \frac{1}{2} pK_a + \frac{1}{2} \lg C_{NaAc} \\ C'_{HCl} = \frac{C_{NaAc} * V_{HCl}}{V + V_{HCl}} = \frac{0.01 * 2}{0.41} \\ pH = 7 - 1.31 + 2.38 + 0.16 = 8.23 \\ \end{array}$$

[1] Ripin, D. H.; Evans, D. A. (4 November 2005). "pKa Table" (PDF). Archived from the original (PDF) on 22 July 2015. Retrieved 19 July 2015.

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