Calculate the pH of a solution made by mixing 50.0 mL of 0.10 M NaOH and 25.0 mL of 0.10 M HClO.?

Question

(Ka= 3.5 × 10-8 for HClO.)

a. ) 1.18

b. ) 1.48

c. ) 11.73

d. ) 12.52

e. ) 12.82

Answer 1

ok first we determine the mol of the base present NaOH which is a strong base # mol = conc x volume =

                                            0.1 M x 0.05 L =  5 x 10^-3 mol

now we have to dermine the conc of H+ resulting from the HClO dissociation

inital mole of HClO = C x  V = 0.025 x  0.1 = 2.5 x 10^-3

the conc is 2.5 x 10^-3  \  V     (total volume is 50 + 25 ml = 0.075 L )  so conc of HClO- is 0.033

HClO ----->  H+   +  ClO-

the initial conc of H+ and ClO- are 0

if HClO- decrease by x then H+ & ClO- increase by x

at equilibrium conc of HClO = (0.033  - X ) H+ = X and ClO- =X

ka =conc of:  H+ x ClO- \ HClO- =  X2 \ 0.033-X = 3.5 x 10^-8

we can determine X   X= 3.398 x 10^-5  M of H+

mol of H+ = 3.398 x 10^-5 \  0.075 L = 4.53 x 10 ^-4 mol

we also calculated the # of mol of the OH- = 5 x 10^-3 mol which will react with the H+ to form H2O  and HO- remains

5 x 10^-3 mol of OH-  - 4.53 x 10 ^-4 mol  =  4.546 x 10^-3 mol HO-  

to finally determine the pH conc of HO- = 4.546 x 10^-3 \ 0.075 L = 0.061 M now pH = 14 + log(conc OH-) = 12.78 which makes answer e the most suitable