Calculate the pH of a buffer solution prepared by dissolving 0.12 mole of cyanic acid, HCNO and 0.65 mole of sodium cyanate, NaCNO in enough water to make 1.0 liter of solution. [Ka(HCNO) = 2.0 × 10-4]

I know that this question has something to do with turning moles into molarity or something, but I am at a loss on how to do it.

2.78

To calculate the pH of a buffer solution, we need to consider the dissociation of the weak acid (HCNO) and the salt (NaCNO) in the solution.

The dissociation of HCNO can be represented as follows:
HCNO ⇌ H+ + CNO-

The dissociation constant (Ka) of HCNO is given as 2.0 × 10^-4. This means that the ratio of the concentration of H+ to the concentration of HCNO is given by the square root of Ka.

First, let's find the initial concentrations of HCNO and CNO- in the solution:
HCNO (initial concentration) = moles of HCNO / volume of solution
HCNO (initial concentration) = 0.12 mol / 1.0 L = 0.12 M

CNO- (initial concentration) = moles of CNO- / volume of solution
CNO- (initial concentration) = 0.65 mol / 1.0 L = 0.65 M

Now, we can calculate the concentration of H+ using the given Ka value:
[H+] = sqrt(Ka) × [HCNO (initial concentration)]
[H+] = sqrt(2.0 × 10^-4) × (0.12 M)
[H+] = (1.414 × 10^-2) × (0.12 M)
[H+] = 1.697 × 10^-3 M

Finally, we can calculate the pH of the buffer solution using the concentration of H+:
pH = -log[H+]
pH = -log(1.697 × 10^-3)
pH ≈ 2.769

Therefore, the pH of the buffer solution is approximately 2.769.