Q.74. The pH a solution obtained by mixing 5 mL of 0.1 M NH₄OH solution with 250 mL of 0.1 M NH₄Cl solution is _______ . (Nearest integer) Given : pK (NH₄OH) = 4.74 log 2 = 0.30 log 3 = 0.48 log 5 = 0.70

Sol. On mixing final volume = 255 ml [NH₄OH] = $\frac{5 \times 0.1}{255}$, [NH₄Cl] = $\frac{250 \times 0.1}{255}$ pOH = pK$_{b}$ + log $\frac{[NH₄Cl]}{[NH₄OH]}$ pOH = 4.74 + log $\frac{250 \times 0.1}{5 \times 0.1}$ pOH = 14 – pOH = 14 – 6.44

What is the correct answer for this exam Chemistry Q74?

The correct answer is 756. Sol. On mixing final volume = 255 ml [NH₄OH] = $\frac{5 \times 0.1}{255}$, [NH₄Cl] = $\frac{250 \times 0.1}{255}$ pOH = pK$_{b}$ + log $\frac{[NH₄Cl]}{[NH₄OH]}$ pOH = 4.74 + log $\frac{250 \times 0.1}{5 \times 0.1}$ pOH = 14 – pOH = 14 – 6.44

Q.74. The pH a solution obtained by mixing 5 mL of 0.1 M NH₄OH solution with 250 mL of 0.1 M NH₄Cl solution is _______ . (Nearest integer) Given : pK (NH₄OH) = 4.74 log 2 = 0.30 log 3 = 0.48 log 5 =…

The correct answer is 756. Sol. On mixing final volume = 255 ml [NH₄OH] = $\frac{5 \times 0.1}{255}$, [NH₄Cl] = $\frac{250 \times 0.1}{255}$ pOH = pK$_{b}$ + log $\frac{[NH₄Cl]}{[NH₄OH]}$ pOH = 4.74 + log $\frac{250 \times 0.1}{5 \times 0.1}$ pOH = 14 – pOH = 14 – 6.44

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Q.74. The pH a solution obtained by mixing 5 mL of 0.1 M NH₄OH solution with 250 mL of 0.1 M NH₄Cl solution is _______ $\times 10^{-2}$. (Nearest integer) Given : pK$_{b}$ (NH₄OH) = 4.74 log 2 = 0.30 log 3 = 0.48 log 5 = 0.70

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