Question

Determine the ph of a 500.0 ml solution of 353.0mg nacn (ka = 2.1 x 10-9 for hcn).

Answer 1

First, find the concentration of the NaCN solution in terms of molarity. The molar mass of NaCN is 49.0072 g/mol

Molarity = mol solute ÷ L solution

Molarity = (353 mg * (1 g / 1000 mg) * (1 / 49.0072 g/mol)) ÷ (500 mL * (1 L / 1000 mL))

Molarity = 0.0144 M NaCN

NaCN is a salt from a weak acid HCN and a strong base NaOH. The reaction is as follows:

HCN + NaOH --> NaCN + H2O

Upon hydrolysis, it becomes:

`CN^(-)` +
`H_(2)O` --> HCN +
`OH^(-)`

Initial: Amount of
`CN^(-)` is 0.0144 M,
`H_(2)O` is in excess, and the products are 0.

Change: This is unknown. Let this be x. For the
`CN^(-)`, it would be –x, while it would be +x for the products HCN and
`OH^(-)`.

Excess: This would be Initial + Change. Thus for
`CN^(-)`, it would be 0.0144 – x. For both products, it would be x for each.

The hydrolysis constant (
`K_(H)`) would be the amount of products over the reactants. Thus,

`K_(H)` = (HCN)*(
`OH^(-)`) / (
`CN^(-)`) = x2 / (0.0144 –x) --> equation 1

`K_(H)` is also equal to the equilibrium constant of water (
`K_(W)` = 1 x 10^-14) divided by
`K_(A)`.

`K_(H)` = 1 x
`10^(-14)` ÷ 2.1 x
`10^(-9)` --> equation 2

Substituting equation 2 to equation 1:

`x^(2)` ÷ (0.0144 –x) = 4.762 x
`10^(-6)`

Solving for x,

X = 2.59 x
`10^(-4)`

As mentioned before, x denotes the products HCN and
`OH^(-)`. Thus the amount of
`OH^(-)` is 2.59 x
`10^(-4)`

To find pH,

pH = 14-(-log
`OH^(-)`)

pH = 14 – (-log 2.59 x
`10^(-4)`)

pH = 10.41

Thus, the answer is 10.41.