Question

a. Calculate the hydronium (H₃O⁺) & hydroxide (OH^-) concentration in a solution prepared by diluting 0.100 mol of HCN to 1L with water. Ka HCN=7.2x10-10.

b. Calculate the hyrdonium (H₃O⁺) & hydroxide (OH^-) concentration in a solution prepared by diluting 0.100 mol NaCN to 1.00L with water. NaCN is a strong electrolyte.

c. what is the pH of a solution prepared by adding 50.0 mL of 0.100M NaOH to 200mL of 0.200 M HCN

Answer 1

A. concentration of HCN = 0.1/1 = 0.1 M

pH of weak acid = 1/2(pka-log C)

pka = -log Ka = -log(7.2*10^(-10)) = 9.14

= 1/2(9.14-log0.1)

pH = 5.07

[H+] = 10^(-pH) = 10^(-5.07) = 8.51*10^-6 M

[OH-] = (1*10^(-14))/(8.51*10^(-6)) = 1.175*10^-9 M

B. NaCN is a salt of weakacid,strongbase.

pH = 7+1/2(pka+log C)

pka of HCN = 9.14

C = concentration of salt = 0.1/1 = 0.1 M

pH = 7+1/2(9.14+log0.1)

   = 11.07

[H+] = 10^(-pH) = 10^(-11.07) = 8.51*10^-12 M

[OH-] = (1*10^(-14))/(8.51*10^(-12)) = 0.001175 M

c.

pH of solution = pka +log(salt/acid)

   = 9.14+log(

No of mol of NaOH = 50/1000*0.1 = 0.005 mol

No of mol of HCN = 20/1000*0.2 = 0.004 mol

so that

concentration of excess NaOH after mixing = (0.005-0.004)/0.07 = 0.0143 M

pOH = -log(OH-) = -log(0.0143) = 1.844

pH = 14-1.844 = 12.156