Calculate the ph of a solution prepared by mixing 40.0 ml of a 0.02 m hcl solution with 200.0 ml of 0.20 m hcn solution. assume volumes to be additive ka for hcn=1.0x10^-10

First, we have to get moles of each HCl & HCN

moles of HCl = volume * molarity         
                  
                       = 0.04 L * 0.02 M = 0.0008 moles

moles of HCN = volume * molarity 

                         = 0.2 L * 0.2 M = 0.04moles 

∴Total moles = 0.04 + 0.0008 = 0.0408 moles

and the total volume = 0.04 L + 0.2 L = 0.24 L

∴the total concentration = total moles / total volume
                                      
                                         = 0.0408 moles / 0.24 L 
                                         = 0.17 M
when we assume [H+] = X

∴ Ka = X^2 / (0.17-X)

by substitution 

∴ 1 x 10^-10 = X^2 / (0.17 -X)

∴X = 4.12 x 10^-6

∴[H+] = 4.12 x 10^-6

∴PH = -㏒(4.12 x 10^-6)
        = 5.4

The answer is !