SATHEE: Equilibrium - Result Question 45

Equilibrium - Result Question 45

####47. Equal volumes of three acid solutions of $p H 3, 4$ and 5 are mixed in a vessel. What will be the $H^{+}$ ion concentration in the mixture?

[2008]

(a) $1.11 \times 10^{- 4} M$

(b) $3.7 \times 10^{- 4} M$

(c) $3.7 \times 10^{- 3} M$

(d) $1.11 \times 10^{- 3} M$

Show Answer

Solution:

(b) $[H_{3} O]^{+}$ for a solution having $p H = 3$ is given by

$[H_{3} O]^{+} = 1 \times 10^{- 3}$ moles/litre $[∴ [H_{3} O]^{+} = 10^{- p H}]$

Similarly for solution having $p H = 4$ , $[H_{3} O]^{+} = 1 \times 10^{- 4} m o l e s /$ litre and for $p H = 5$

$[H_{3} O^{+}] = 1 \times 10^{- 5} m o l e s /$ litre

Let the volume of each solution in mixture be 1L, then total volume of mixture solution

$= (1 + 1 + 1) L = 3 L$

Total $[H_{3} O]^{+}$ ion present in mixture solution $= (10^{- 3} + 10^{- 4} + 10^{- 5})$ moles

Then $[H_{3} O]^{+}$ ion concentration of mixture solution

$\begin{aligned} = \frac{10^{- 3} + 10^{- 4} + 10^{- 5}}{3} M = \frac{0.00111}{3} M \\ = 0.00037 M = 3.7 \times 10^{- 4} M . \end{aligned}$

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