Question

17.65 mL of a 0.110 M sodium hydroxide solution is needed to titrate 25.00 mL of a hydrochloric acid solution to the equivalence point.
A. Write the balanced equation for the neutralization reaction.
B. How many moles of sodium hydroxide are used for the titration?.
C. How many moles of hydrochloric acid reacted with sodium hydroxide?
D. What is the molar concentration (Molarity) of the hydrochloric acid solution?

Answer 1

Answer on the question #54073 – Chemistry – General chemistry

Question:

17.65 mL of a 0.110 M sodium hydroxide solution is needed to titrate 25.00 mL of a hydrochloric acid solution to the equivalence point.

A. Write the balanced equation for the neutralization reaction.

B. How many moles of sodium hydroxide are used for the titration?

C. How many moles of hydrochloric acid reacted with sodium hydroxide?

D. What is the molar concentration (Molarity) of the hydrochloric acid solution?

Answer:

A. Reaction equation:

$$NaOH + HCl = NaCl + H_2O.$$

B. The number of the moles of sodium hydroxide:

$$n(NaOH) = cV = 0.110 \cdot 17.65 \cdot 10^{-3} = 1.9415 \cdot 10^{-3} \text{ mol}.$$

C. The number of the moles of hydrochloric acid, reacted with sodium hydroxide, according to the equation, is equal to the number of the moles of sodium hydroxide, added:

$$n(HCl) = n(NaOH) = 1.9415 \cdot 10^{-3} \text{ mol}.$$

D. The molar concentration of hydrochloric acid solution is:

$$c(HCl) = \frac{n(HCl)}{V(HCl)} = \frac{1.9415 \cdot 10^{-3}}{25 \cdot 10^{-3}} = 0.07766 \text{ mol } L^{-1}.$$

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