Calculate the pH of the resultant mixtures:
10ml of 0.2M Ca(OH)2 + 25 mL of 0.1 M HCI
Answers
Explanation:
Total number of moles present in 10 mL of 0.2 M calcium hydroxide are 100010×0.2=0.002 moles.
Total number of moles present in 25 mL of 0.1 M HCl are 100025×2=0.0025 moles.
Ca(OH)2+2HCl→CaCl2+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 0.002−0.00125=0.00075 moles.
Total volume of the solution is 10+25=35 mL.
The molarity of the solution is 350.00075×1000=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 H2SO4 is 100010×0.01=0.0001 mol.
Total number of moles present in 10 mL of 0.01 M Ca(OH)2=100010×0.01=0.0001 mole.
Ca(OH)2+H2SO4→CaSO