Back to Directory
NEET CHEMISTRYMedium

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

1.11×10−4 M1.11 \times 10^{-4} \text{ M}

B

3.7×10−4 M3.7 \times 10^{-4} \text{ M}

C

3.7×10−3 M3.7 \times 10^{-3} \text{ M}

D

1.11×10−3 M1.11 \times 10^{-3} \text{ M}

Step-by-Step Solution

The concentration of H+H^+ ions can be calculated from the pH values: [H+]=10−pH[H^+] = 10^{-pH} For the first solution (pH = 3): [H+]1=10−3 M[H^+]_1 = 10^{-3}\text{ M} For the second solution (pH = 4): [H+]2=10−4 M[H^+]_2 = 10^{-4}\text{ M} For the third solution (pH = 5): [H+]3=10−5 M[H^+]_3 = 10^{-5}\text{ M} Let the volume of each solution be VV. Total moles of H+H^+ in the mixture = (10−3+10−4+10−5)×V=(100×10−5+10×10−5+1×10−5)×V=111×10−5×V=1.11×10−3×V(10^{-3} + 10^{-4} + 10^{-5}) \times V = (100 \times 10^{-5} + 10 \times 10^{-5} + 1 \times 10^{-5}) \times V = 111 \times 10^{-5} \times V = 1.11 \times 10^{-3} \times V. Total volume of the mixture = V+V+V=3VV + V + V = 3V. The final H+H^+ ion concentration = Total moles of H+Total volume=1.11×10−3×V3V=0.37×10−3 M=3.7×10−4 M\frac{\text{Total moles of } H^+}{\text{Total volume}} = \frac{1.11 \times 10^{-3} \times V}{3V} = 0.37 \times 10^{-3}\text{ M} = 3.7 \times 10^{-4}\text{ M}.

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started