What is the pH of a 0.710 M solution of NaCN (Ka of HCN is 4.9 × 10⁻¹⁰)?

Question

Please Answer, im struggling. What is the pH of a 0.710 M solution of NaCN (Ka of HCN is 4.9 × 10⁻¹⁰)?

Answer 1

11.6

The reaction is: CN- + H2O = HCN + OH-

Note that Na in NaCN is a spectator ion and doesn't need to be

included in this equation.

We need to find the value of Kb, the concentration of OH-, [OH-] to find pOH, which is -log[OH-], in order to plug it into the equation pH+pOH=14 to solve for solution pH.

Ka​×Kb​=Kw​=1.0×10^(−14)

Kb=10^(−14)/Ka​=1.0×10^(−14)/4.9×10^(−10)=2.04*10^(-5)

Kb= [BH+][OH-]/[B]

In this case, Kb=[HCN][OH-]/[CN-]

The concentrations of HCN and OH are both equal in this reaction due to their equal coefficients in the equation above.

We already know [CN-]=0.710M equal to [NaCN] given.

So, Kb=[HCN][OH-]/[CN-]=2.04*10^(-5) -->

[HCN][OH-]/0.710=2.04*10^(-5)

[HCN][OH-]=1.4484*10^(-5)

As said earlier, we know the [HCN]=[OH-].

So, we can say [OH-]^2=√1.4484*10^(-5) --> [OH-]=√1.4484*10^(-5) (the square root of 2.04*10^(-5)) --> [OH-]=3.8058*10^(-3) Remember not to round early!

Converting that to pOH, we have -log[OH-]=-log(3.8058*10^(-3))=2.419554

Using the equation pH+pOH=14, we have pH+2.419554=14, giving us a pH of 11.580446; 11.6 with respect to significant figures

This high pH indicates a basic solution which is expected due to the ion's incredibly low Ka that indicates how much it is lacking in acidity.