Chemistry, asked by vihan2162023, 6 hours ago

Calculate the pH after 10.0 mL of 0.125 M NaOH is added to 30.0 mL of 0.105 M HCHO2.

Answers

Answered by raj19414
0

Answer:

(a) Total number of moles present in 10 mL of 0.2 M calcium hydroxide are

1000

10×0.2

=0.002 moles.

Total number of moles present in 25 mL of 0.1 M HCl are

1000

25×2

=0.0025 moles.

Ca(OH)

2

+2HCl→CaCl

2

+2H

2

O

1 mole of calcium hydroxide reacts with 2 moels of HCl.

0.0025 moles of HCl will react with 0.00125 moles of calium hydroxide.

Total number of moles of calcium hydroxide unreacted are 0.002−0.00125=0.00075 moles.

Total volume of the solution is 10+25=35 mL.

The molarity of the solution is

35

0.00075×1000

=0.0214M

[OH

]=2×0.0214=0.0428M

pOH=−log0.0428=1.368

pH=14−pOH=14−1.368=12.635

(b) Total number of moles present in 10 mL of 0.01 M H

2

SO

4

is

1000

10×0.01

=0.0001 mol.

Total number of moles present in 10 mL of 0.01 M Ca(OH)

2

=

1000

10×0.01

=0.0001 mole.

Ca(OH)

2

+H

2

SO

4

→CaSO

4

+2H

2

O

0.0001 mole of Ca(OH)

2

will react completely with 0.0001 mole of H

2

SO

4

.

Hence, the resuting solution is neutral with pH 7.0

(c) Total number of moles in 10 mL of 0.1 M H

2

SO

4

=

1000

10×0.1

=0.001 mole.

Total number of moles present in 10 mL of 0.1 M KOH =

1000

10×0.1

=0.001 mole.

2KOH+H

2

SO

4

→K

2

SO

4

+2H

2

O

2 moles of KOH reacts with 1 mole of H

2

SO

4

.

0.001 mole of KOH will react with 0.0005 mole of H

2

SO

4

.

Number of moles of H

2

SO

4

left =0.001−0.0005=0.005M

Volume of solution is 10+10=20mL

Molarity of the solution is

20

0.0005×1000

=0.025M.

[H

+

]=2×2.5×10

−2

=0.05M

pH=−log[H

+

]=−log0.05=1.3.

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