75 mL of 0.2 M HCl is mixed with 25 mL of 1 M HCl. To this solution, 300 mL of distilled water is added. What is the pH of the resultant solution?

Options

A. 1
B. 2
C. 4
D. 0.2

Answer

A

Explanation

`V_(1) = 75 mL, M_(1) = 0.2 M, V_(2) = 25 mL`
`M_(2) = 1 M, V_(3) = 800 mL`
`V_(3) = V_(1) + V_(2) = 75 + 25 = 100 mL`
`M_(3) = (M_(1)V_(1) + M_(2)V_(2))/V_(1) + V_(2) = (0.2 xx 75 + 1 xx 25)/75 + 25 = 4/100 = 0.4 M`
Total volume, `V^(1) = V_(1) +V_(2)+V_(3) = 75+25+300 = 400 mL`
`M_(3)V_(3) = M^(1)V^(1)`
`0.4 xx 100 = M^(1) xx 400`
`M^(1) = 0.4 xx 100/400`
`M^(1) = 0.1 M`
Concentration of resultant solution of HCl = 0.1 M
`pH = -"log"[H^(+)] = -"log"(0.1) = -"log"(10^(-1)) = -(-1)"log"_(10)10 = 1`