Solveeit Logo
Login

Question

Chemistry Question on Equilibrium

Equal volumes of three acid solutions of pH 3, 4 and 5 are mixed in a vessel. What will be the H+H^+ ion concentration in the mixture?

A

1.11×10−4M1.11 \times 10^{-4}M

B

3.7×10−4M3.7 \times 10^{-4}M

C

3.7×10−3M3.7 \times 10^{-3}M

D

1.11×10−3M1.11 \times 10^{-3}M

Answer

3.7×10−4M3.7 \times 10^{-4}M

Explanation

Solution

Let the volume of each acid = VV
pH of first, second and third acids = 3, 4 and 5 respectively
[H+][H^+] of first acid (M1)=1×10−3  [∴H+=1×10−pH] (M_1) = 1 \times 10^{-3} \, \, [ \therefore H^+ = 1 \times 10^{-pH}]
[H+][H^+] of second acid (M2)=1×10−4(M_2) = 1 \times 10^{-4}
        [H+]\, \, \, \, \, \, \, \, [H^+] of third acid (M3)=1×10−5(M_3) = 1 \times 10^{-5}
Total [H+][H^+] concentrated of mixture
(M)=M1V1+M2V2+M3V3V1+V2+V3(M) = \frac{M_1V_1 + M_2V_2 + M_3V_3}{V_1 + V_2 + V_3 }
      =1×10−3×V+1×10−4×V+1×10−5×VV+V+V\, \, \, \, \, \, = \frac{ 1 \times 10^{-3} \times V + 1 \times 10^{-4} \times V + 1 \times 10^{-5} \times V}{V + V + V}
      =1×10−3×V(1+0.1+0.01)3V\, \, \, \, \, \, = \frac{1 \times 10^{-3} \times V (1 + 0.1 + 0.01) }{3V}
      =1.11×10−33=3.7×10−4M\, \, \, \, \, \, = \frac{1.11 \times 10^{-3} }{3} = 3.7 \times 10^{-4} M