Ionic equilibrium

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Question

100 mL solution of pH = 6 is diluted to 1000 mL. Resulting solution has pH

Difficult

Solution

pH = 6
$∴ [H^{+}] = 10^{- 6} M$
$Also solution is acidic (pH < 7) and dilution will not change its acidic nature.$
${[H^{+}]}_{diluted} = \frac{100 \times 10^{- 6}}{1000} = 1 \times 10^{- 7} M$
[H⁺ ] from H₂O is also required while calculating resultant concentration.
$Total [H^{+}] = {[H^{+}]}_{diluted} + {[H^{+}]}_{H_{2} O} = 1 \times 10^{- 7} + 1 \times 10^{- 7} = 2 \times 10^{- 7} M pH = - \log 2 \times 10^{- 7} = 6.7$

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Ionic equilibrium

Unlock the full solution & master the concept.

100 mL solution of pH = 6 is diluted to 1000 mL. Resulting solution has pH

pH = 6∴ H+=10−6M Also solution is acidic (pH<7) and dilution will not change its acidic nature.H+diluted =100×10−61000 =1×10−7M[H+ ] from H2O is also required while calculating resultant concentration. Total H+=H+diluted +H+H2O =1×10−7+1×10−7 =2×10−7M pH=−log 2×10−7 =6.7

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