Answer:
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
