Millimoles of calcium hydroxyide required to produce 100 mL of the aqueous solution of pH 12 is x × 10 - 1 . The value of x is _____ (Nearest integer). Assume complete dissociation.

We know, pH = - log [ H + ] , pH = - log [ OH - ] , pH + pOH = 14 ∵ pH = 12 ∴ H + = 10 - 12 M ∴ OH - = 10 - 2 M We also know, Ca ( OH ) 2 ⇌ 2 OH - + Ca 2 + ∴ Ca ( OH ) 2 = 5 × 10 - 3 M 5 × 10 - 3 = milli moles of Ca OH 2 100 mL milli moles of Ca OH 2 = 5 × 10 - 1 Ans. = 5

Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumpH of Strong and Weak Acids and BasesMedium2 minPYQ_2023

ChemistryMediumstatement

Millimoles of calcium hydroxyide required to produce $100 mL$ of the aqueous solution of $pH 12$ is $x \times 10^{- 1}$ . The value of $x$ is _____ (Nearest integer).
Assume complete dissociation.

Answer:

Solution:

We know,
$pH = - \log [H^{+}]$ , $pH = - \log [{OH}^{-}], pH + pOH = 14$
$∵ pH = 12$

$∴ [H^{+}] = 10^{- 12} M$

$∴ [{OH}^{-}] = 10^{- 2} M$
We also know,
$Ca (OH)_{2} ⇌ 2 {OH}^{-} + {Ca}^{2 +}$

$∴ [Ca (OH)_{2}] = 5 \times 10^{- 3} M$

$5 \times 10^{- 3} = \frac{milli moles of Ca {(OH)}_{2}}{100 mL}$

milli moles of $Ca {(OH)}_{2} = 5 \times 10^{- 1}$

Ans. $= 5$

Stream:JEESubject:ChemistryTopic:Ionic EquilibriumSubtopic:pH of Strong and Weak Acids and Bases

⏱ 2mℹ️ Source: PYQ_2023

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