Question

A student performs a titration using 1.00 M NaOH to find the unknown molarity of a solution of HCl. The student records the data as shown below. What is the molarity of the solution of MCl?

Base: final burette reading	21.05 mL
Base: initial burette reading	6.05 mL
mL of base used
Acid: final burette reading	44.15 mL
Acid: initial burette reading	14.15 mL
mL of acid used

• A. $0.75M$
• B. $0.50M$
• C. $0.25M$
• D. $0.10M$
• E. $2.00M$

Answer 1

The correct option is B $0.50M$

According to the law of equivalence,

$N_1{V}_1=N_2{V}_2$

$HCl$ and $NaOH$ has normality equal to molarity (Monobasic and monoacidic)

$M_1{V}_1=M_2{V}_2$

Here $M_1=$ Molarity of base $=1.0M$

$V_1=$ volume of base used$=21.05-6.05=15.0ml$

$M_2=$ Molarity of acid$=?$

$V_2=$ Volume of acid used$=44.15-14.15=30ml$

$M_2=\frac{M_1{V}_1}{V_2}=\frac{1\times 15}{30}=0.50M$