Calculate the pH of -C- M N a 2 S solution in terms of pK1- pK2 and logC - K 1- of H 2 S -5-0 - 10-5 K 2 of H 2 S -2-5 - 10-11-A- pH -7- pk 1- pk 2-2-1-2log CB- pH -7- pk 1-2-1-2log CC- pH -7- pk 2-2-1-2log CD- pH -7- pk 1-2 pk 2-1-2log C

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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Chemistry

Calculate the...

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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
$K b = \frac{0.05 [B^{+}]^{- 4}}{0.1 [B O H]} = \frac{0.05 \times 10^{-} 4}{0.1} = 5 \times 10^{- 5} P K b = - l o g 5 + 5 = 4.31$ .

Suggest Corrections

Similar questions

Q. 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}

H_{2} S = 2.5 \times 10^{- 11}

Q. pH of 0.01 M

H S^{-}

will be -

Q. Calculate pH of

1

M

N a H C O_{3}

. Given

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

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

Hence, pH of

0.01 M

N a H_{2} {P O}_{4}

is:

Q. Calculate pH of 0.05M potassium hydrogen phthalate,

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

H_{2} C_{8} H_{4} O_{4} + H_{2} O \to H_{3} O^{+} + H C_{8} H_{4} O_{4}^{-} . p K_{1} = 2.94

H C_{8} H_{4} O_{4}^{-} + H_{2} O \to H_{3} O^{+} + C_{8} H_{4} O_{4}^{- 2} . p K_{2} = 5.44

Join BYJU'S Learning Program

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 $H_{2} S = 2.5 \times 10^{- 11}$ .