Question

Sodium hydroxide is used extensively in acid-base titrations because it is a strong, inexpensive base. A sodium hydroxide solution was standardized by titrating 38.87 mL of 0.1994 M standard hydrochloric acid. The initial buret reading of the sodium hydroxide was 1.16 mL, and the final reading was 36.18 mL. What was the molarity of the base solution?

Answer 1

Answer: The molarity of the base solution is 0.22 M.

Explanation:

According to the neutralization law,

$n_1M_1V_1=n_2M_2V_2$

where,

$M_1$ = molarity of $HCl$ solution = 0.1994 M

$V_1$ = volume of $HCl$ solution = 38.87 ml

$M_2$ = molarity of $NaOH$ solution = ?

$V_2$ = volume of $NaOH$ solution = Final reading - Initial reading = 36.18 - 1.16 = 35.02 ml

$n_1$ = valency of $HCl$ = 1

$n_2$ = valency of $NaOH$ = 1

$1\times 0.1994\times 38.87=1\times M_2\times 35.02$

$M_2=0.22M$

Therefore, the molarity of the base solution is 0.22 M.

Answer 2

Answer:

0.221M

Explanation:

applying molarity equation