Verified Solution • 2025–26 Session moderate

QJEEM22Chcb95: JEE Main 2022 Chemistry - Class 11 JEEM22Chcb95 Solution: of is mixed with of . The of the mixture is _________. Given:

200 mL200 \mathrm{~mL} of 0.01 M HCl0.01 \,\mathrm{M} \,\mathrm{HCl} is mixed with 400 mL400 \mathrm{~mL} of 0.01 M H2SO40.01 \,\mathrm{M} \,\mathrm{H}_{2} \mathrm{SO}_{4} . The pH\mathrm{pH} of the mixture is _________.

Given: log⁡2=0.30,log⁡3=0.48,log⁡5=0.70,log⁡7=0.84,log⁡11=1.04\log {2}=0.30, \log 3=0.48, \log 5=0.70, \log 7=0.84, \log 11=1.04

Multiple Choice Options

  • A 1.14
  • B 1.78
  • C 2.34
  • D 3.02

Step-by-Step Verified Solution

[H+]amp;=0.01×200+2×0.01×400600amp;=0.01+2×0.01×23amp;=0.01+0.043amp;=53×10−2pHamp;=−log⁡[H+]amp;=−log⁡(53×10−2)amp;=−[log⁡53+log⁡10−2]amp;=−[log⁡5−log⁡3−2]amp;=−0.7+0.48+2amp;=2.48−0.7amp;=1.78\begin{aligned} {\left[\mathrm{H}^{+}\right] } &=\frac{0.01 \times 200+2 \times 0.01 \times 400}{600} \\ &=\frac{0.01+2 \times 0.01 \times 2}{3} \\ &=\frac{0.01+0.04}{3} \\ &=\frac{5}{3} \times 10^{-2} \\ \mathrm{pH} &=-\log \left[\mathrm{H}^{+}\right] \\ &=-\log \left(\frac{5}{3} \times 10^{-2}\right) \\ &=-\left[\log \frac{5}{3}+\log 10^{-2}\right] \\ &=-[\log 5-\log 3-2] \\ &=-0.7+0.48+2 \\ &=2.48-0.7 \\ &=1.78 \end{aligned}

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