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:
[tex] CN^{-} [/tex] + [tex] H_{2}O [/tex] --> HCN + [tex] OH^{-} [/tex]
Initial: Amount of [tex] CN^{-} [/tex] is 0.0144 M, [tex] H_{2}O [/tex] is in excess, and the products are 0.
Change: This is unknown. Let this be x. For the [tex] CN^{-} [/tex], it would be –x, while it would be +x for the products HCN and [tex] OH^{-} [/tex].
Excess: This would be Initial + Change. Thus for [tex] CN^{-} [/tex], it would be 0.0144 – x. For both products, it would be x for each.
The hydrolysis constant ([tex] K_{H} [/tex]) would be the amount of products over the reactants. Thus,
[tex] K_{H} [/tex] = (HCN)*([tex] OH^{-} [/tex]) / ([tex] CN^{-} [/tex]) = x2 / (0.0144 –x) --> equation 1
[tex] K_{H} [/tex] is also equal to the equilibrium constant of water ([tex] K_{W} [/tex] = 1 x 10^-14) divided by [tex] K_{A} [/tex].
[tex] K_{H} [/tex] = 1 x [tex] 10^{-14} [/tex] ÷ 2.1 x [tex] 10^{-9} [/tex] --> equation 2
Substituting equation 2 to equation 1:
[tex] x^{2} [/tex] ÷ (0.0144 –x) = 4.762 x [tex] 10^{-6} [/tex]
Solving for x,
X = 2.59 x [tex] 10^{-4} [/tex]
As mentioned before, x denotes the products HCN and [tex] OH^{-} [/tex]. Thus the amount of [tex] OH^{-} [/tex] is 2.59 x [tex] 10^{-4} [/tex]
To find pH,
pH = 14-(-log [tex] OH^{-} [/tex])
pH = 14 – (-log 2.59 x [tex] 10^{-4} [/tex])
pH = 10.41
Thus, the answer is 10.41.
