a) 50 ml of 0.1 M HCl + 50 ml of 0.01 M HCl
b) 50 ml of 0.1 M HCl + 50 ml of 0.1 M H2SO4
c) 50 ml of 0.1 M HCl + 50 ml of 0.1 M CH3COOH (Ka=1.8x10-5)
d) 50 ml of 0.1 M HCl + 20 ml of 0.1 M NaOH
e) 50 ml of 0.1 M HCl + 50 ml of 0.1 M NaOH
f) 20 ml of 0.1 M HCl + 50 ml of 0.1 M NaOH
g) 50 ml of 0.1 M HCN(Ka=10-5) + 50 ml 0.1 M CH3COOH(Ka=1.8x10-5)
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3 Answers
301) Millimoles of 0.1M HCl = 5
Millimoles of 0.01M HCl=0.5
Total millimoles=5.5
[H+]= 5.5/100
pH= 2 - 0.74 = 1.26
2) Millimoles of HCl=5
Millimoles of H2SO4 = 10
Total millimoles of H+=15
Concentration = 15/100 =
pH= 2 - 1.176= 0.824
3) Millimoles of HCl = 5
[H+]= 5/100
pH=2-0.7=1.3
4) After neutralisation,
Millimoles of HCl left = 3
[H+] = 3/70
pH= 1.84
5) Clearly, pH = 7
306) After neutralisation,
Millimoles of NaOH left = 3
[OH-]=3/70
pOH=1.84
pH = 14-1.84 = 12.16
7) Millimoles of HCN, CH3COOH = 5
[H+]=√(Ka1C1+Ka2C2)
=√(1.8x10-5x5/100 + 10-5x5/100)
pH= 4-1.072 = 2.928
8) After neutralisation,
Millimoles of HCl left = 2.5
[H+]= 2.5/75=1/30
pH = 1.477