1. pH = ?
V1 = 600mL = 0,6 L NaOH
C1 = 0,005M
V2 = 400mL = 0,4L HCl
C2 = 0,005M
C1 = n1/V1 => n1 = C1*V1 = 0,005*0,6 = 0,003 moli NaOH
C2 = n2/V2 => n2 = C2*V2 = 0,005*0,4 = 0,002 moli HCl
n NaOH exces = 0,001 moli
V final = 1000mL = 1L
C final = n NaOH exces/V final = 0,001/1 = 10⁻³ M
[OH⁻] = C final = 10^(-pOH) = 10⁻³M => pOH = 3
pH = 11
2. pH = ?
V = 500mL
m = 1,825g HCl
C = md/M*Vs = 1,825/36,5*0,5 = 0,1M
HCl -> A.T => [HCl] = C = [H₃O⁺] = 10^(-pH) = 10⁻¹M => pH = 1