100. mLmL of 0.200 MM HClHCl is titrated with 0.250 MM NaOHNaOH. Part A What is the pH of the solution after 50.0 mLmL of base has been added

Explanation:

100 mL of .2M HCl = 100 x .2 mL of M HCl = 20 mL of M HCl

50 mL of .25M NaOH = 50 x .25 mL of M NaOH = 12.5 mL of M NaOH

HCl remaining after titration = 20 - 12.5 mL of M HCl

= 7.5 mL of M HCl

volume of solution = 100 + 50 = 150 mL

concentration of HCl = C

7.5 x M = 150 x C

C = .05 M .

[ H⁺ ] = .05 = 5 x 10⁻²

pH = - log [ 5 x 10⁻² ]

= 2 - log 5

= 2 - .7

1.3

pH = 1.3