Question

a 0.100 m solution of naoh is used to titrate an hcl solution of unknown concentration. to neutralize the solution, an average volume of the titrant was 38.2 ml. the starting volume of the hcl solution was 20 ml. whats the concentration of the hcl?
a) 3.34 m
b) 0.191 m
c) 0.284 m
d) 0.788 m

Explanation:

Step1: Recall the titration formula

For a strong acid - strong base titration (HCl and NaOH), the formula is \(M_1V_1 = M_2V_2\), where \(M_1\) is the molarity of HCl, \(V_1\) is the volume of HCl, \(M_2\) is the molarity of NaOH, and \(V_2\) is the volume of NaOH.

Step2: Identify the known values

We know that \(M_2 = 0.100\space M\), \(V_2 = 38.2\space mL\), and \(V_1 = 20\space mL\). We need to find \(M_1\).

Step3: Rearrange the formula to solve for \(M_1\)

From \(M_1V_1 = M_2V_2\), we can rearrange it to \(M_1=\frac{M_2V_2}{V_1}\)

Step4: Substitute the known values into the formula

Substitute \(M_2 = 0.100\space M\), \(V_2 = 38.2\space mL\) and \(V_1 = 20\space mL\) into the formula:
\(M_1=\frac{0.100\space M\times38.2\space mL}{20\space mL}\)
First, calculate the numerator: \(0.100\times38.2 = 3.82\)
Then, divide by the denominator: \(\frac{3.82}{20}= 0.191\space M\)

Answer:

B) 0.191 M