WBJEE 2025 Physics and Chemistry Papers

Since both solutions are strong acids, we assume complete dissociation. The hydrogen ion concentration in each solution is given by: $[\mathrm{H}^{+}] =10^{- \mathrm{pH}} .$ For Solution A $\mathrm{pH}=6.0 ):$ $[\mathrm{H}^{+}] _A=10^{-6} \mathrm{M}$ For Solution B $( \mathrm{pH}=4.0 ):$ $[\mathrm{H}^{+}] _B=10^{-4} \mathrm{M} .$ When equal volumes of the two solutions are mixed, the total hydrogen ion concentration in the new solution is the average of the two:$[\mathrm{H}^{+}]_{\;\text{mix}\;} = \frac{[\mathrm{H}^{+}]_A + [\mathrm{H}^{+}]_B}{2} = \frac{10^{-6} + 10^{-4}}{2} .$ Calculate the numerator: $10^{-6}+10^{-4}=0.000001+0.0001 \approx 0.000101 .$ Then divide by 2 : $[\mathrm{H}^{+}]_{\text{mix}} \approx \frac{0.000101}{2} \approx 5.05 \times 10^{-5} \ \mathrm{M} .$ $[\mathrm{H}^{+}]_{\;\text{mix}\;} \approx \frac{0.000101}{2} \approx 5.05 \times 10^{-5} \ \mathrm{M} .$ Now, we convert this concentration back to pH : $\mathrm{pH}_{\text{mix}} =-\log _{10} (5.05 \times 10^{-5}) .$ Breaking the logarithm into parts: Since $\log _{10}(5.05) \approx 0.7$, we get:$\mathrm{pH}_{\;\text{mix}\;} \approx 5-0.7=4.3 .$Thus, the pH of the new solution is approximately 4.3 , which lies between 4 and 5 .Therefore, the correct answer is:Option C: between 4 and 5.