Calculate the pH of a solution formed by mixing equal volumes of two solutions A and B of a strong acid having pH = 6 and pH = 4 respectively

<p>This is a Short Answer Type Questions as classified in NCERT Exemplar</p><p><strong>Ans: we can find the pH of the solution as-</strong></p><p>pH of solution&nbsp;A=6</p><p>Hence, concentration of&nbsp;^[H⁺]&nbsp;ion in solution&nbsp;A=10-6mol&nbsp;L^-1</p><p>pH of solution&nbsp;B=4</p><p>Therefore, concentration of&nbsp; [<em>H</em>]⁺&nbsp;ion in solution&nbsp;B=10^-4mol&nbsp;L^-1</p><p>On mixing one litre of each solution, total volume =&nbsp;1L+1L=2L.</p><p>Amount of&nbsp;H⁺&nbsp;ions in&nbsp;1L&nbsp;of solution</p><p>&nbsp;A= concentration× Volume (V)= 10^-6mol× IL&nbsp;</p><p>Amount of&nbsp;H^+&nbsp;ions in&nbsp;1L&nbsp;of solution&nbsp;B=10^-4mol×1L</p><p>∴&nbsp;Total amount of&nbsp;H⁺&nbsp;ions in the solution formed by mixing solutions&nbsp;A&nbsp;and&nbsp;B&nbsp;is&nbsp; (10-5mol+10-4mol)</p><p>This amount is present in&nbsp;2L&nbsp;solution, </p><p>∴&nbsp;Total&nbsp;H⁺</p><p>=210^-4 (1+0.01)? =21.01×10-4? mol&nbsp;L-1</p><p>=21.01×10^-4? mol&nbsp;L^-1&nbsp;=0.5×10⁻⁴mol&nbsp;L−1&nbsp;=5×10^-5mol&nbsp;L^-1.</p><p>pH=−log [<em>H</em>]⁺=−log (5×10^-5)=−<em>log</em>5+ (−5<em>log</em>10)</p><p>= −log5+5=5−log5=5−0.6990=4.3010=4.3</p><p>Thus, the pH of the solution will be&nbsp;4.3.</p>

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