300 mL of 5-10-2M HCl is allowed to react with 200 mL of 5-10-2M NaOH at 300K- The pH of the resulting solution at 300K is

The number of moles of hydrogen ions present in HCl solution =0.3 L × 0.05 mol/L = 0.015 mol The number of moles of hydroxide ions present in NaOH solu ...

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Standard XII

Chemistry

Salt of Strong Acid and Strong Bases

300 mL of 5...

Question

$300$ mL of $5 \times 10^{- 2} M$ $H C l$ is allowed to react with $200$ mL of $5 \times 10^{- 2} M$ $N a O H$ at $300 K$ . The $p H$ of the resulting solution at $300 K$ is

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Solution

The number of moles of hydrogen ions present in HCl solution
$= 0.3 L \times 0.05 m o l / L = 0.015 m o l$
The number of moles of hydroxide ions present in NaOH solution
$= 0.2 L \times 0.05 m o l / L = 0.010 m o l$
The number of moles of hydrogen ions remaining after neutralization
$0.015 m o l - 0.010 m o l = 0.005 m o l$
Total volume $= 0.5 L$
$[H^{+}] = \frac{0.005 m o l}{0.5 L} = 0.01 M$
$p H = - l o g [H^{+}]$
$= - l o g 0.01$
$= 2$

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300 mL of 5×10−2M HCl is allowed to react with 200 mL of 5×10−2M NaOH at 300K. The pH of the resulting solution at 300K is

$300$ mL of $5 \times 10^{- 2} M$ $H C l$ is allowed to react with $200$ mL of $5 \times 10^{- 2} M$ $N a O H$ at $300 K$ . The $p H$ of the resulting solution at $300 K$ is