Chemical Equilibrium

20 mL of 0.1 M NH₄Oh is mixed with 40 mL of 0.05 M HCl. The pH of the mixture is nearest to: (Given K_b (NH₄OH) = 1 * 10^-5, log2 = 0.30, log3 = 0.48, log5 = 0.69, log7 = 0.84, log11 = 1.04

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Contributor-Level 10

$[N H_{4} C l] = \frac{2}{60} = \frac{1}{30} M$

$p H = 7 - \frac{1}{2} P K_{b} - \frac{1}{2} l o g C$

$= 7 - \frac{5}{2} - \frac{1}{2} l o g (\frac{1}{30}) = 5.24$

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7 months ago

Ka for butyric acid (C₃H₇COOH) is 2 * 10^-⁵. The pH of 0.2 M solution of butyric acid is ____________* 10^-¹ (Nearest integer)

[Given log2 = 0.30]

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Contributor-Level 10

Ka for C₃H₇COOH = 2 * 10^-⁵

$p K a = - l o g (2 * 1 0^{- 5}) = 5 - l o g 2$

=5 – 0.3 = 4.7

pH of 0.2 (M) solution =

$p H = \frac{p K a - l o g C}{2}$

$= \frac{1}{2} (4.7) - \frac{1}{2} l o g (0.2)$

$p H = 27 * 1 0^{- 1}$

Ans 27

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7 months ago

Value of K_p for the equilibrium reaction N₂O_4(g) $\overset{}{?}$ 2NO_2(g) at 288 K is 47.9. The K_c for this reaction at same temperature is_________. (Nearest integer)

(R = 0.083 L bar K^-1 mol^-1)

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Contributor-Level 10

K_p = 47.9 T = 288 K

K_C =? R = 0.083 L bar/K mol

$N_{2} O_{4} (g) ? 2 N O_{2} (g)$

Using

$K_{p} = K_{C} {(R T)}^{Δ n g}$

Here ;

$Δ n g = 1$

49.7 = K_C (0.083 * 288)¹

K_C = 2

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7 months ago

A solution is 0.1M in Cl^- and 0.001M in $C r O_{4}^{2 -}$ . Solid AgNo₃ is gradually added to it. Assuming that the addition does not change in volume and K_sp(AgCl) = 1.7 * 10^-10M² and K_sp (Ag₂CrO₄) = 1.9 * 10^-12M³

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Vishal Baghel

Contributor-Level 10

$[C l^{-}] = 1 0^{- 1} M$

$[C r {O_{4}}^{- -}] = 1 0^{- 3} M$

for AgCl ppt¹, ${[A g^{+}]}_{r e q} = \frac{1.7 * 1 0^{- 10}}{1 0^{- 1}} M$

For Ag₂CrO₄ppt², ${[A g^{+}]}_{r e q} = \sqrt[]{\frac{1.9 * 1 0^{- 12}}{1 0^{- 3}}} M$

${[A g^{+}]}_{r e q} = 4.3 * 1 0^{- 5} M$

Being lower concentration of [Ag⁺] in case of AgCl, it will precipitate first.

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7 months ago

If 80g of copper sulphate CuSO₄ $\cdot$ 5H₂O is dissolved in deionized water to make 5L of solution. The concentration of the copper sulphate solution is x * 10^-3 mol L^-1. The value of x is________.

[Atomic masses Cu : 63.54u, S : 32u, O : 16u, H : 1u]

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Contributor-Level 10

Mass of CuSO₄. 5H₂O = 80 g

Volume of solution = 5L

Molar mass of CuSO₄.5H₂O = 249.54g/ml

$C o n c e n t r a t i o n = \frac{m o l e s}{v o l u m e o f s o l u t i o n}$

$\begin{array}{l} = \frac{0.32}{5} = 0.064 M \\ = 64 * 1 0^{- 3} M \end{array}$

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7 months ago

The pH of a solution obtained by mixing 50 mL of 1 M HCl and 30 mL of 1 M NaOH is x * 10^-4. The value of x is ________. (Nearest integer)

[log 2.5 = 0.3979]

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Vishal Baghel

Contributor-Level 10

50 ml of 1 (M) HCl + 30 ml of 1 (M) NaOH

NaOH + HCl -> NaCl + H₂O

30 * 1 mmol 50 * 1 mmol

0 mmol 20 mmol

${[H^{+}]}_{m i x} = \frac{20}{50 + 30} M = \frac{20}{80} M = \frac{1}{4} M = 0.25 M$

$x * 1 0^{- 4} = 6021 * 1 0^{- 4}$

$∴$ x = 6021

Ans. = 6021

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7 months ago

When 5.1g of solid NH₄HS is introduced into a two litre evacuated flask at 27°C, 20% of the solid decomposes into gaseous ammonia and hydrogen sulphide. The K_p for the reaction at 27°C is x * 10^-2. The value of x is_______. (Integer answer)

[Given R = 0.082 L atm K^-1 mol^-1]

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Contributor-Level 10

Moles of NH₄HS initially taken = $\frac{5.1}{51} = 0.1 m o l$

$N H_{4} H S (S) + N H_{3} (g) + H_{2} S (g)$

t = $\infty$ 0 0.1 mol 0 0

t = $\infty$ 0.1 (1 - 0.2) 0.1 * 0.2 0.1 * 0.2

$P_{N H_{3}} = \frac{n R T}{V} = \frac{0.1 * 0.2 * 0.082 * 300}{2} = 0.246 a t m = P_{H_{2} S}$

$K_{P} = P_{N H_{3}} * P_{H_{2} S} = {(0.246)}^{2} = 0.060516 = 6.05 * 1 0^{- 2}$

x = 6 (Nearest integer).

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7 months ago

The neutralization occurs when 10 mL of 0.1 M acid 'A' is allowed to react with 30 mL of 0.05 M base M base M(OH)₂. The basicity of the acid 'A' is___________.

[M is a metal]

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Raj Pandey

Contributor-Level 9

Acid = M (OH)2 Salt + H2O

M.E of Acid = me of Base

$∴ 10 * (0.1 * n f) = (0.05 * 2) * 30$

$∴ n_{f} = 3$

Thus basicity of acid = 3

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8 months ago

The OH^- concentration in a mixture of 5.0 mL of 0.0504 M NH₄Cl and 2 mL of 0.0210 M NH₃ solution is x * 10^-6 M. The value of x is________. (Nearest integer)

[Given K_w = 1 * 10^-14 and K_b = 1.8 * 10^-5]

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Vishal Baghel

Contributor-Level 10

0.0504 M NH₄Cl of 5ml => millimole of $N H_{4}^{+} = 0.0504 * 5$

0.0210 M NH₃ of 2ml => millimole of NH₃ = 0.0210 * 2

It is a basic buffer.

Total volume = 7ml

$H e r e, K_{b} = \frac{[O H^{-}] * [N H_{4}^{+}]}{[N H_{4} O H]}$

$∴ 1.8 * 1 0^{- 5} = \frac{[O H^{-}] * 0.0504 * 5}{0.0210 * 2}$

$∴ [O H^{-}] = \frac{1.8 * 1 0^{- 5} * 0.0210 * 2}{0.0504 * 5} = 0.3 * 1 0^{- 5} M$

$∴ [O H^{-}] = 3 * 1 0^{- 6} M$

$∴ x = 3$

Ans. = 3

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8 months ago

The equilibrium constant K_c at 298 K for the reaction

A + B $⇌$ C + D

is 100. Starting with an equimolar solution with concentrations of A, B, C and D all equal to 1M, the equilibrium concentration of D is_________* 10^-2 M. (Nearest integer)

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