HA + BOH ⇌ BA + H 2 O ; K c = P × 1 0 − Q (in correct scientific notation). The ( P + Q ) value is:

HA + BOH ⇌ BA + H 2 O K c = K w K a ⋅ K b = 1 0 − 14 3 × 1 0 − 7 × 3 × 1 0 − 7 = 9 = 9 × 1 0 − 0

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ChemistryChemical EquilibriumProblems involving KcMedium2 minQB

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Passage / Comprehension

HA is a weak acid, and BOH is a weak base. The ionization constant ( $K_{a}$ ) of HA is $3 \times 1 0^{- 7}$ , and the ionization constant ( $K_{b}$ ) of BOH is $3 \times 1 0^{- 7}$ at $298 K$ .

$HA + BOH ⇌ BA + H_{2} O$ ; $K_{c} = P \times 1 0^{- Q}$ (in correct scientific notation). The $(P + Q)$ value is:

Answer:

9.00

Solution:

$HA + BOH ⇌ BA + H_{2} O$

$K_{c} = \frac{K _{a} \cdot K _{b}}{K _{w}} = \frac{3 \times 1 0 ^{- 7} \times 3 \times 1 0 ^{- 7}}{1 0 ^{- 14}} = 9 = 9 \times 1 0^{- 0}$

Stream:JEESubject:ChemistryTopic:Chemical EquilibriumSubtopic:Problems involving Kc

⏱ 2mℹ️ Source: QB

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