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Ionic Equilibrium question

2021 · 31 Aug · Shift 2 · Q17
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Ionic Equilibrium question

2021 · 31 Aug · Shift 2 · Q17

JEE MainChemistryIonic EquilibriumNumerical+4 / −1
The pH of a solution obtained by mixing 50 mL of 1 M HCl and 30 mL of 1 M NaOH is x ×\times× 10 −-− 4. The value of x is ‾\underline{\hspace{2cm}}​. (Nearest integer) [log 2.5 = 0.3979]
Numerical answer
View written solutionFree

Correct answer: 6021

  1. Write the neutralization reaction

HCl+NaOH→NaCl+H2O\mathrm{HCl + NaOH \rightarrow NaCl + H_2O}HCl+NaOH→NaCl+H2​O

Both are strong electrolytes, so we first calculate moles.

  1. Moles of HCl and NaOH

For HCl: nHCl=1×501000=0.05 moln_{\mathrm{HCl}} = 1 \times \frac{50}{1000} = 0.05\text{ mol}nHCl​=1×100050​=0.05 mol

For NaOH: nNaOH=1×301000=0.03 moln_{\mathrm{NaOH}} = 1 \times \frac{30}{1000} = 0.03\text{ mol}nNaOH​=1×100030​=0.03 mol

  1. Find excess acid after neutralization

NaOH is limiting, so it neutralizes equal moles of HCl.

Excess HCl: 0.05−0.03=0.02 mol0.05 - 0.03 = 0.02\text{ mol}0.05−0.03=0.02 mol

  1. Total volume after mixing

Vtotal=50+30=80 mL=0.08 LV_{\text{total}} = 50 + 30 = 80\text{ mL} = 0.08\text{ L}Vtotal​=50+30=80 mL=0.08 L

  1. **Concentration of excess } H^+ $$

Since HCl is strong, [H+]=0.020.08=0.25 M[H^+] = \frac{0.02}{0.08} = 0.25\text{ M}[H+]=0.080.02​=0.25 M

  1. Calculate pH

pH=−log⁡[H+]=−log⁡(0.25)\mathrm{pH} = -\log [H^+] = -\log(0.25)pH=−log[H+]=−log(0.25)

Now, 0.25=2.5×10−10.25 = 2.5 \times 10^{-1}0.25=2.5×10−1

So, log⁡(0.25)=log⁡(2.5)−1=0.3979−1=−0.6021\log(0.25) = \log(2.5) - 1 = 0.3979 - 1 = -0.6021log(0.25)=log(2.5)−1=0.3979−1=−0.6021

Hence, pH=−(−0.6021)=0.6021\mathrm{pH} = -(-0.6021) = 0.6021pH=−(−0.6021)=0.6021

The question says pH is x×10−4x \times 10^{-4}x×10−4.

Thus, x×10−4=0.6021x \times 10^{-4} = 0.6021x×10−4=0.6021

x=0.6021×104=6021x = 0.6021 \times 10^4 = 6021x=0.6021×104=6021

  1. Nearest integer

x=6021x = 6021x=6021

Comparison with stored answer: Stored correct answer is 6021, which matches our result.

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