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Question

Calculate the pH of the resultant mixtures:
a) 10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl
b) 10 mL of 0.01M
H2SO4+ 10 mL of 0.01M Ca(OH)2
c) 10 mL of 0.1M
H2SO4 + 10 mL of 0.1M KOH

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Solution

(a) Total number of moles present in 10 mL of 0.2 M calcium hydroxide are 10×0.21000=0.002 moles.
Total number of moles present in 25 mL of 0.1 M HCl are 25×21000=0.0025 moles.
Ca(OH)2+2HClCaCl2+2H2O
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 0.0020.00125=0.00075 moles.
Total volume of the solution is 10+25=35 mL.
The molarity of the solution is 0.00075×100035=0.0214M
[OH]=2×0.0214=0.0428M
pOH=log0.0428=1.368
pH=14pOH=141.368=12.635
(b) Total number of moles present in 10 mL of 0.01 M H2SO4 is 10×0.011000=0.0001 mol.
Total number of moles present in 10 mL of 0.01 M Ca(OH)2=10×0.011000=0.0001 mole.
Ca(OH)2+H2SO4CaSO4+2H2O
0.0001 mole of Ca(OH)2 will react completely with 0.0001 mole of H2SO4.
Hence, the resuting solution is neutral with pH 7.0
(c) Total number of moles in 10 mL of 0.1 M H2SO4=10×0.11000=0.001 mole.
Total number of moles present in 10 mL of 0.1 M KOH =10×0.11000=0.001 mole.
2KOH+H2SO4K2SO4+2H2O
2 moles of KOH reacts with 1 mole of H2SO4.
0.001 mole of KOH will react with 0.0005 mole of H2SO4.
Number of moles of H2SO4 left =0.0010.0005=0.005M
Volume of solution is 10+10=20mL
Molarity of the solution is 0.0005×100020=0.025M.
[H+]=2×2.5×102=0.05M
pH=log[H+]=log0.05=1.3.

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