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Standard IX

Chemistry

pH of a Solution

50mL of 0.1...

Question

$50 m L$ of $0.1 M N a O H$ is added to $60 m L$ of $0.15 M H_{3} P O_{4}$ solution ( $K_{1}, K_{2}$ , and $K_{3}$ amy be taken as $10^{3}, 10^{4}$ , and $10^{12}$ respectively), the $p H$ of the mixture would be about:

3.1

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5.5

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4.1

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Solution

The correct option is A $3.1$
$H^{+} = \frac{N_{1} V_{1} - N_{2} V_{2}}{V_{1} + V_{2}}$ , $(-)$ sign as the acid and base are given.
$H^{+} = \frac{.1 \times 50 - .3 \times 60}{110} = .02909$
As the $P H = - l o g H^{+}$
After taking the log both side,
$P H \approx 3.1$

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Similar questions

50

mL of

0.1 M N a O H

is added to 60 mL of 0.15 M

H_{3} P O_{4}

solution . The

p H

of the mixture would be about:

(K_{1}, K_{2}

and

K_{3}

for

H_{3} P O_{4}

are

10^{- 3}, 10^{- 8}

and

10^{- 13}

respectively)

(l o g 2 = 0.3)

50 m L

0.2 M H C l

is titrated against the

0.1 M N a O H

solution. What will be the

p H

of the solution when

20 m L N a O H

is added ?

Q. If

K_{1}, K_{2} a n d K_{3}

are equilibrium constant for formation of

A D, A D_{2} a n d A D_{3}

respectively as follow

A + D ⇌ A D, A D + D ⇌ A D_{2}, A D_{2} + D ⇌ A D_{3}

then equilibrium constant 'K" for

A + 3 D ⇌ A D_{3}

is related as

For $C u^{2 +}$ ion in $[C u (H_{2} O)_{4} (e n)]^{2 +}, [C u (H_{2} O)_{2} (e n)_{2}]^{2 +}$ and $[C u (e n)_{3}]^{2 +}$ with the stability constant $k_{1}, k_{2}$ and $k_{3}$ respectively, which of the following holds true?

Q. How much

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

must be added to one litre of

0.005 M

solution of

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

in order to make

1 L

of the solution of

p H = 6.7

K_{1} = 7.1 \times 10^{- 3}

;

K_{2} = 6.3 \times 10^{- 8}

;

K_{3} = 4.5 \times 10^{- 13}

for

H_{3} {P O}_{4}

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50mL of 0.1 M NaOH is added to 60mL of 0.15 M H3PO4 solution (K1,K2, and K3 amy be taken as 103,104, and 1012 respectively), the pH of the mixture would be about:

The correct option is A 3.1H+=N1V1−N2V2V1+V2, (−) sign as the acid and base are given.H+=.1×50−.3×60110=.02909As the PH=−logH+After taking the log both side,PH≈3.1

$50 m L$ of $0.1 M N a O H$ is added to $60 m L$ of $0.15 M H_{3} P O_{4}$ solution ( $K_{1}, K_{2}$ , and $K_{3}$ amy be taken as $10^{3}, 10^{4}$ , and $10^{12}$ respectively), the $p H$ of the mixture would be about:

The correct option is A $3.1$
$H^{+} = \frac{N_{1} V_{1} - N_{2} V_{2}}{V_{1} + V_{2}}$ , $(-)$ sign as the acid and base are given.
$H^{+} = \frac{.1 \times 50 - .3 \times 60}{110} = .02909$
As the $P H = - l o g H^{+}$
After taking the log both side,
$P H \approx 3.1$