For hydrocyanic acid, HCN, Ka = 4.9 × 10-10. Calculate the pH of 0.20 M NaCN. What is the concentration of HCN in the solution? for the ph i got 10 is that right?

by using the ICE table :

initial 0.2 M 0 0
change -X + X +X
Equ (0.2 -X) X X

when Ka = (X) (X) / (0.2-X)
so by substitution:

4.9x10^-10 = X^2 / (0.2-X) by solving this equation for X
∴X ≈ 10^-6
∴[HCN] = 10^-6

and PH = -㏒[H+]
= -㏒ 10^-6
= 6

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Alleei Alleei · Answer 2 · 2019-10-16T11:22:31+02:00

Answer : The concentration of HCN in the solution is [tex]2.0\times 10^{-3}M[/tex]

Explanation :

First we have to calculate the value of [tex]pK_a[/tex].

The expression used for the calculation of [tex]pK_a[/tex] is,

[tex]pK_a=-\log (K_a)[/tex]

Now put the value of [tex]K_a[/tex] in this expression, we get:

[tex]pK_a=-\log (4.9\times 10^{-10})[/tex]

[tex]pK_a=10-\log (4.9)[/tex]

[tex]pK_a=9.3[/tex]

Now we have to calculate the pH of the solution.

The hydrolysis reaction will be,

[tex]NaCN+H_2O\rightarrow HCN+NaOH[/tex]

Formula used :

[tex]pH=7+\frac{1}{2}[pKa+\log C][/tex]

[tex]pH=7+\frac{1}{2}[9.3+\log (0.20)][/tex]

[tex]pH=11.3[/tex]

Now we have to calculate the concentration of HCN in the solution.

Using Henderson Hesselbach equation :

[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]

[tex]pH=pK_a+\log \frac{[NaCN]}{[HCN]}[/tex]

Now put all the given values in this expression, we get:

[tex]11.3=9.3+\log (\frac{0.20}{[HCN]})[/tex]

[tex][HCN]=2.0\times 10^{-3}M[/tex]

Therefore, the concentration of HCN in the solution is [tex]2.0\times 10^{-3}M[/tex]