Question

A 20.0-mL sample of 0.30 M HClO was titrated with 0.30 M NaOH. The following data were collected during the titration. Determine Ka for HClO. Data Provided: mL NaOH added pH 5.00 6.98 10.00 7.46 15.00 7.93 20.00 10.31 (Hint: Find the volume of NaOH at equivalence point)

Answer 1

Answer:

$K_a=3.46 \times 10^{-8}.$

Explanation:

We know at equivalent point,

$M_1V_1=M_2V_2\\M_{acid}V_{acid}=M_{base}V_{base}\\0.30 \times20=0.30 \times V_{base}\\V_{base}=20 ml.$

So, equivalent point is obtained at 20 ml.

Therefore, half equivalent point is obtained at 10 ml.

Also , at 10 ml NaOH , pH= 7.46 .

We know,

$pH=pK_a$

Also, $pK_a=-\ln K_a$

Taking anti log both sides .

We get $K_a=3.46 \times 10^{-8}.$

Hence, it is the required question.