a solution was prepared by mixing 40 cm3 of 0.02M HNO3 solution with 60 cm3 of 0.03M KOH solution Kw=1x10-14\
Calculate the concentration of excess H+ or OH ‾ ions left in solution?
please help me
KOH reacts with HNO3 :
KOH + HNO3 → KNO3 + H2O
1mol KOH reacts with 1 mol HNO3
Mol HNO3 in 40cm³ of 0.02M solution = 40/1000*0.02 = 0.0008 mol HNO3
Mol KOH in 60cm³ of 0.03M solution = 60/1000*0.03 = 0.0018 mol KOH
These react and 0.0018 - 0.0008 = 0.001 mol KOH remains dissolved in 40+60 = 100 cm³ solution
Molarity of KOH in solution = 0.001/0.1 = 0.01M KOH solution
Because KOH is a strong base ,
[OH-] = 0.01M First answer
Concentration of H+ ions
[H+] [OH-] = 10^-14
[H+] = 10^-14/[OH-]
[H+] = 10^-14 / 0.01
[H+] = 1*10^-12M second answer.
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Verified answer
KOH reacts with HNO3 :
KOH + HNO3 → KNO3 + H2O
1mol KOH reacts with 1 mol HNO3
Mol HNO3 in 40cm³ of 0.02M solution = 40/1000*0.02 = 0.0008 mol HNO3
Mol KOH in 60cm³ of 0.03M solution = 60/1000*0.03 = 0.0018 mol KOH
These react and 0.0018 - 0.0008 = 0.001 mol KOH remains dissolved in 40+60 = 100 cm³ solution
Molarity of KOH in solution = 0.001/0.1 = 0.01M KOH solution
Because KOH is a strong base ,
[OH-] = 0.01M First answer
Concentration of H+ ions
[H+] [OH-] = 10^-14
[H+] = 10^-14/[OH-]
[H+] = 10^-14 / 0.01
[H+] = 1*10^-12M second answer.