Question

What is the ph of a solution prepared by mixing 25.00 ml of 0.10 m ch3co2h with 25.00 ml of 0.020 m ch3co2na? assume that the volume of the solutions are additive and that k a = 1.8 × 10-5 for ch3co2h?

Answer 1

Answer:- The pH of the buffer solution will be 4.04 .

Solution:- Sodium acetate added to acetic acid makes a buffer solution and the pH of buffer solution is easily calculated by using Handerson equation:

$pH=pKa+log(\frac{base}{acid})$

We are adding 25.00 mL of each, so the total volume of the buffer solution will be 50.00 mL.

We can calculate the concentration of base which is sodium acetate and acid by using dilution equation:

$M_1V_1=M_2V_2$

$M_2=\frac{M_1V_1}{V_2}$

for base:

$M_2=\frac{0.020*25.00}{50.00}$

$M_2=0.010M$

Similarly, for acid:

$M_2=\frac{0.10*25.00}{50.00}$

$M_2=0.050M$

pKa is calculated from given Ka value as:

$pKa=-logKa$

$pKa=-log(1.8*10^-^5)$

pKa = 4.74

Let's plug in the values in Handerson equation:

$pH=4.74+log(\frac{0.010}{0.05})$

pH = 4.74 - 0.70

pH = 4.04