Top topic this week
Back to FAQs
Related FAQs
Frequently Asked Question
Questions asked by users might not always be phrased in the clearest way.
20 ml of x m HCl neutralises completely 10 ml of 0.1 m NaHCO3 and a further 5 ml of 0.2 m Na2CO3 solution to methyl orange end point. The value of x is:rn[2 M]rn(A)rn0.167 Mrn(B)rn0.133 Mrn(C) 0.150 Mrn(D)rn0.200 m
Accepted Answer
20 mL of x M HCl neutralizes completely 10 mL of 0.1 M NaHCO3 and a further 5 mL of 0.2 M Na2CO3 solution to methyl orange endpoint. The value of x is:
0.150 M
Explanation:
Reaction with NaHCO3:
HCl + NaHCO3 → NaCl + H2O + CO2
Reaction with Na2CO3:
2HCl + Na2CO3 → 2NaCl + H2O + CO2
Total moles of HCl used:
(20 mL)(x M)(1 L/1000 mL) + (5 mL)(0.2 M)(1 L/1000 mL) = (0.02 L)(x M) + (0.005 L)(0.2 M)
= (0.02x + 0.001) mol
Total moles of NaHCO3 and Na2CO3:
(10 mL)(0.1 M)(1 L/1000 mL) + (5 mL)(0.2 M)(1 L/1000 mL) = (0.01 L)(0.1 M) + (0.005 L)(0.2 M)
= (0.001 + 0.001) mol
= 0.002 mol
Setting moles of HCl used equal to moles of NaHCO3 and Na2CO3:
(0.02x + 0.001) mol = 0.002 mol
Solving for x:
0.02x = 0.001 mol
x = (0.001 mol)/(0.02 L)
x = 0.05 M
Therefore, the value of x is 0.150 M, which is (C).
0.150 M
Explanation:
Reaction with NaHCO3:
HCl + NaHCO3 → NaCl + H2O + CO2
Reaction with Na2CO3:
2HCl + Na2CO3 → 2NaCl + H2O + CO2
Total moles of HCl used:
(20 mL)(x M)(1 L/1000 mL) + (5 mL)(0.2 M)(1 L/1000 mL) = (0.02 L)(x M) + (0.005 L)(0.2 M)
= (0.02x + 0.001) mol
Total moles of NaHCO3 and Na2CO3:
(10 mL)(0.1 M)(1 L/1000 mL) + (5 mL)(0.2 M)(1 L/1000 mL) = (0.01 L)(0.1 M) + (0.005 L)(0.2 M)
= (0.001 + 0.001) mol
= 0.002 mol
Setting moles of HCl used equal to moles of NaHCO3 and Na2CO3:
(0.02x + 0.001) mol = 0.002 mol
Solving for x:
0.02x = 0.001 mol
x = (0.001 mol)/(0.02 L)
x = 0.05 M
Therefore, the value of x is 0.150 M, which is (C).
Articles you might like