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The correct option is At equivalence point, Moles of BOH = Moles of HCl (Reacted) c × 20 = 12 × 0.3 c=0.18 (OH^ ions from BOH is neutralized by H^+ ions fro ...

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Chemistry

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Question

Calculate the pH of 'C' M $N a_{2} S$ solution in terms of pK1, pK2 and logC . [K1 of $H_{2} S = 5.0 \times 10^{- 5} K_{2}$ of $H_{2} S = 2.5 \times 10^{- 11}$ .

p H = \frac{7 + (p k 1 + p k 2)}{2} + \frac{1}{2} l o g C

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p H = \frac{7 + (p k 1)}{2} + \frac{1}{2} l o g C

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p H = \frac{7 + (p k 2)}{2} + \frac{1}{2} l o g C

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p H = \frac{7 + (p k 1)}{2 (p k 2)} + \frac{1}{2} l o g C

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Solution

At equivalence point,
Moles of BOH = Moles of HCl (Reacted)
$c \times 20 = 12 \times 0.3$
$c = 0.18$
( $O H^{-}$ ions from BOH is neutralized by $H^{+}$ ions from acid)
Now, consider the date pH of the base solution was 10 upon addition of 4mL 0.3 HCL solution
$[O H] = 10^{- 4} M$
$B O H + H C l ⇌ B C l + H_{2} O$
New concentration of BOH $= \frac{0.18 \times 20}{24} = 0.15$
$B O H_{(a q)} + H C l_{(a q)} ⇌ B C l + H_{2} O$
initial 0.15 0.15 0 0
Final 0.10 0 0.05
$K b = \frac{0.05 [B^{+}]^{- 4}}{0.1 [B O H]} = \frac{0.05 \times 10^{-} 4}{0.1} = 5 \times 10^{- 5} P K b = - l o g 5 + 5 = 4.31$ .

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p K_{1} = 6.38

H C O_{3}^{-} + H_{2} O ⇌ C O_{3}^{2 -} + H_{3} O^{+};

p K_{2} = 10.26

H_{2} O + H_{3} {P O}_{4} ⇌ H_{3} O^{+} + H_{2} {P O}_{4}^{-}; p K_{1} = 2.15

H_{2} O + H_{2} {P O}_{4}^{-} ⇌ H_{3} O^{+} + H_{2} {P O}_{4}^{2 -}; p K_{2} = 7.20

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Q. Calculate pH of 0.05M potassium hydrogen phthalate,

K H C_{8} H_{4} O_{4}

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Calculate the pH of 'C' M Na2S solution in terms of pK1, pK2 and logC . [K1 of H2S=5.0×10−5K2 of H2S=2.5×10−11.

Calculate the pH of 'C' M $N a_{2} S$ solution in terms of pK1, pK2 and logC . [K1 of $H_{2} S = 5.0 \times 10^{- 5} K_{2}$ of $H_{2} S = 2.5 \times 10^{- 11}$ .