Question

What is the pH of a buffer made from 0.130 mol of HCNO (Ka = 3.5 × 10⁻⁴) and 0.410 mol of NaCNO in 2.0 L of solution?

Answer 1

Answer:

$pH=3.95$

Explanation:

Hello,

In this case, by using the Henderson-Hasselbach equation we can compute the pH as follows:

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

Whereas the concentration of the acid (HCNO) and the base (NaCNO) are computed by considering the given volume:

$[base]=\frac{0.410mol}{2.0L}=0.205M$

$[acid]=\frac{0.130mol}{2.0L}=0.065M$

Therefore, the pH turns out:

$pH=-log(3.5x10^{-4})+log(\frac{0.205M}{0.065M} )\\\\pH=3.95$

Regards.