What is the ph of a 0.023 m solution of hcn (ka = 4.9 x 10-10)?

The ionic equation is simply:

HCN --> H+  +  CN-

Therefore the formula for Ka is:

Ka = [H+] [CN-] / [HCN] 
[H+] = [CN-] = x
4.9 x 10^-10 = x^2 / 0.023
x^2 = 1.127 x 10^-11 
x = 3.357 x 10^-6 = [H+] = [CN-]

pH = -log [H+] 
pH = -log 3.357 x 10^-6 
pH = 5.47 

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole