Question

Q: A 25.5 mL aliquot of HCl (aq) of unknown concentration was titrated with 0.113 M NaOH (aq). It took 51.2 mL of the base to reach the endpoint of the titration. The concentration (M) of the acid was __________.

A: 0.227

Please help explain how to get the answer.

Answer 1

M*25.5=.113*51.2

Answer 2

simply equate the equilvalence

Answer 3

To solve, you need to discern the acid:base ratio for the titration. The acid HCl contains one mol of electrons per mol of molecules. The base NaOH contains one mol of protons per mol of molecules. Therefore, your ratios are equal and no balancing need occur. This gives you a ratio of (1)HCl : (1) NaOH. This will be referred to as: \[M _{1} : M _{2}\] where the compounds are associated respectively.

The equation to determine molarity is as follows: \[M = m/L\] where m is the number of moles and L is the measurement in liters.

As the mol ratio for this specific problem is 1 : 1, M1 will equal M2.
Plugging in to the molarity equation, M1 is equal to: \[M _{1} = m/.0255\]
The molarity of the second compound is given as 0.113. Therefore, one can use the measurement in liters and the molarity to determine the number of moles in the solution.
\[m = (0.0512) \times (0.113) = .0057856\]
This can then be plugged into the equation for M1 to yield the answer of the concentration of the acid: \[M _{1} = (.0057856) / (.0255)\] =0.227