Question

59.1 mL of 0.877 M nitric acid is added to 25.7 mL of sodium hydroxide, and the resulting solution is found to be acidic. 17.9 mL of 0.955 M barium hydroxide is required to reach neutrality. What is the molarity of the original sodium hydroxide solution?

a) 0.95 M
b) 1.01 M
c) 0.92 M
d) 1.10 M

Answer 1

To find the molarity of the original sodium hydroxide solution, we can use the balanced chemical equation for the reaction between nitric acid (HNO3) and sodium hydroxide (NaOH). We can then calculate the moles of NaOH used in the neutralization reaction and determine the molarity of the original sodium hydroxide solution.

Step-by-step explanation:

To find the molarity of the original sodium hydroxide solution, we can use the balanced chemical equation for the reaction between nitric acid (HNO3) and sodium hydroxide (NaOH):

2HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l)

From the given information, we know that 59.1 mL of 0.877 M nitric acid solution is added to 25.7 mL of sodium hydroxide (NaOH). We also know that 17.9 mL of 0.955 M barium hydroxide (Ba(OH)2) is required to reach neutralization. Since the mole ratio between NaOH and Ba(OH)2 is 1:2 in the neutralization reaction, we can calculate the moles of NaOH used with the following equation:

moles of NaOH = 2 * (moles of Ba(OH)2)

By substituting all the known values into the equation and solving for the moles of Ba(OH)2, we can determine the moles of NaOH used and then calculate the molarity of the original sodium hydroxide solution.