Question

Calculate the pH of 720. mL of a 0.425-M solution of hydrocyanic acid before and after the addition of 0.249 mol of sodium cyanide.

Answer 1

Answer:

The pH of hydrocyanic acid before and after the addition of sodium cyanide is 0.371 and 9.121

Explanation:

Given: Concentration of hydrocyanic acid (HCN) = 0.425 M

To find: pH

Reaction:

$HCN \,\,\,\,\,\rightarrow \,\,\,\,\,\,H^+ + CN^-$

0.425M     0.425M

$pH= -[logH^+]\\pH=-[log\,0.425]\\pH= 0.371$

pH after adding 0.249 mol of sodium cyanide (NaCN)

Pka of HCN= 9.21

Concentration of sodium cyanide = $\frac{0.249}{0.720}\,M$

Concentration of NaCN= 0.346 M

According to Henderson–Hasselbalch equation:

$pH= p_k_a + log\frac{salt}{acid}$

Putting the values in the above equation,

$pH= 9.21 + log\frac{0.346}{0.425} \\\\pH= 9.21 - 0.089\\pH= 9.121$

Note: Pka value of HCN is assumed as 9.21

Learn more about Henderson–Hasselbalch equation;

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