53.9 mL of 0.991 M nitric acid is added to 45.3 mL of sodium hydroxide, and the resulting solution is found to be acidic.
20.5 mL of 1.51 M potassium hydroxide is required to reach neutrality.
What is the molarity of the original sodium hydroxide solution?
Total mols HNO3 (too much) = M x L = ?
- mols KOH added to correct = M x L =
= mols HNO3 used to titrate the NaOH.
Then M NaOH = mols NaOH/L NaOH.
The answer is appox 0.5 M
To find the molarity of the original sodium hydroxide solution, we can use the concept of neutralization reaction.
First, let's determine the amount of nitric acid (HNO3) that reacted with sodium hydroxide (NaOH) in the first reaction.
The balanced chemical equation for the reaction between nitric acid and sodium hydroxide is:
HNO3 + NaOH → NaNO3 + H2O
From the balanced equation, we can see that 1 mole of nitric acid reacts with 1 mole of sodium hydroxide to produce 1 mole of sodium nitrate and 1 mole of water.
Given that the volume of nitric acid is 53.9 mL and its molarity is 0.991 M, we can calculate the number of moles of nitric acid as follows:
moles of HNO3 = volume (L) × molarity (mol/L)
= 53.9 mL ÷ 1000 mL/L × 0.991 mol/L
= 0.05336 moles
Since the balanced chemical equation shows a 1:1 mole ratio between nitric acid and sodium hydroxide, we can conclude that the amount of sodium hydroxide that reacted is also 0.05336 moles.
Next, let's determine the molarity of the sodium hydroxide solution.
The balanced chemical equation for the reaction between sodium hydroxide and potassium hydroxide is:
NaOH + KOH → KNOH + H2O
From the balanced equation, we can see that 1 mole of sodium hydroxide reacts with 1 mole of potassium hydroxide to produce 1 mole of potassium nitrate and 1 mole of water.
Given that the volume of potassium hydroxide is 20.5 mL and its molarity is 1.51 M, we can calculate the number of moles of potassium hydroxide as follows:
moles of KOH = volume (L) × molarity (mol/L)
= 20.5 mL ÷ 1000 mL/L × 1.51 mol/L
= 0.031055 moles
Since the balanced chemical equation shows a 1:1 mole ratio between sodium hydroxide and potassium hydroxide, we can conclude that the amount of sodium hydroxide that reacted is also 0.031055 moles.
Now, we can relate the amount of sodium hydroxide that reacted in the two reactions:
0.05336 moles (from the first reaction) = 0.031055 moles (from the second reaction)
To find the molarity of the original sodium hydroxide solution, we divide the number of moles of sodium hydroxide by the total volume of the solution (in liters) before the reactions occurred:
molarity of NaOH = moles of NaOH / volume of NaOH solution (L)
Since the volumes of both nitric acid and potassium hydroxide solutions are given, we can calculate the volume of the original sodium hydroxide solution:
volume of NaOH solution = volume of NaOH added - volume of KOH added
= 45.3 mL - 20.5 mL
= 24.8 mL
Now we can calculate the molarity of the original sodium hydroxide solution:
molarity of NaOH = 0.031055 moles / 24.8 mL ÷ 1000 mL/L
= 1.25 M
Therefore, the molarity of the original sodium hydroxide solution is 1.25 M.