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What volumes of HCl was added if 20mL of 1M NaOH is titrated with 1M HCl to produce a pH=2?
the answer is 20.4.
the answer is 20.4.
need help solving this gchem question
- Thread starter skim1988
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Suppose you are adding 20mL of 1M HCl to 20mL of 1M NaOH and yield a solution with pH=7 (strong acid+strong base = neutralization) and a volume of 40 mL .
In order to make a solution of pH=2, you now have to add x mL of 1M HCl more to the 40 mL solution.
Using M1V1 = M2V2 (moles of HCl added must equal to moles of HCl in the final solution),
1M * (x mL) = 0.01M * (40 mL + x mL)
0.01 on the right side of the equation merely represents molarity of HCl in the final solution. Since the final pH must be 2, [H+] = 10^(-2)=0.01M. If the hydrogen ion concentration is 0.01M , then there would naturally be 0.01 M of HCl. And 40 mL is derived from the initial solution volume of 20 mL of HCl and 20 mL of NaOH.
Algebraically solving the above equation yields the following:
x = 0.4 + 0.01 * x
0.99x = 0.4
Thus, x is approximately 0.4, meaning 0.4 mL of 1M HCl should be added in addition to 20 mL of 1M HCl that was already added. This yields total of 20.4 mL of 1M HCl being added to 20 mL of 1M NaOH.
PS. In this case, converting mL to L is unnecessary during M1V1 = M2V2 calculation as the units will "cancel" out in the end.
In order to make a solution of pH=2, you now have to add x mL of 1M HCl more to the 40 mL solution.
Using M1V1 = M2V2 (moles of HCl added must equal to moles of HCl in the final solution),
1M * (x mL) = 0.01M * (40 mL + x mL)
0.01 on the right side of the equation merely represents molarity of HCl in the final solution. Since the final pH must be 2, [H+] = 10^(-2)=0.01M. If the hydrogen ion concentration is 0.01M , then there would naturally be 0.01 M of HCl. And 40 mL is derived from the initial solution volume of 20 mL of HCl and 20 mL of NaOH.
Algebraically solving the above equation yields the following:
x = 0.4 + 0.01 * x
0.99x = 0.4
Thus, x is approximately 0.4, meaning 0.4 mL of 1M HCl should be added in addition to 20 mL of 1M HCl that was already added. This yields total of 20.4 mL of 1M HCl being added to 20 mL of 1M NaOH.
PS. In this case, converting mL to L is unnecessary during M1V1 = M2V2 calculation as the units will "cancel" out in the end.
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