1) What volume of 1.25 mol/L hydrobromic acid, HBr(aq) will neutralize 75.0 ml of 0.895 mol/L magnesium hydroxide Mg(OH)2(aq)?
2) a 25.00 mL sample of a nitric acid solution , HNO3(aq) is neutralized by 18.55mL of a 0.1750 mol/L sodium hydroxide, NaOH(aq) what is the concentration of the nitric acid solution?
Please help me with these two questions. Your help is very much appreciated! Thanks!!!:)
I have answered this question above.
Sure, I can help you with these questions! Let's go through each one step by step.
Question 1:
To determine the volume of the hydrobromic acid needed to neutralize the given amount of magnesium hydroxide, we can use the concept of stoichiometry. The balanced chemical equation for the neutralization reaction is:
2HBr(aq) + Mg(OH)2(aq) -> MgBr2(aq) + 2H2O(l)
From the balanced equation, we can see that 2 moles of hydrobromic acid react with 1 mole of magnesium hydroxide.
First, let's calculate the number of moles of magnesium hydroxide in the given 75.0 mL of 0.895 mol/L solution:
moles of Mg(OH)2 = concentration (mol/L) x volume (L)
= 0.895 mol/L x 0.0750 L
= 0.0671 mol
Since the stoichiometry ratio between HBr and Mg(OH)2 is 2:1, we need twice as many moles of hydrobromic acid to fully react with the magnesium hydroxide.
moles of HBr needed = 2 x moles of Mg(OH)2
= 2 x 0.0671 mol
= 0.134 mol
Now, to find the volume of 1.25 mol/L hydrobromic acid needed, we can use the formula:
volume (L) = moles / concentration (mol/L)
= 0.134 mol / 1.25 mol/L
= 0.1072 L
The volume of 1.25 mol/L hydrobromic acid required to neutralize 75.0 mL of 0.895 mol/L magnesium hydroxide is 0.1072 L (or 107.2 mL).
Question 2:
In this question, we want to find the concentration of the nitric acid solution.
First, let's determine the number of moles of sodium hydroxide used to neutralize the nitric acid. From the balanced chemical equation:
HNO3(aq) + NaOH(aq) -> NaNO3(aq) + H2O(l)
The stoichiometry ratio between HNO3 and NaOH is 1:1. Therefore, the moles of HNO3 are equal to the moles of NaOH used in the reaction.
moles of NaOH = concentration (mol/L) x volume (L)
= 0.1750 mol/L x 0.01855 L
= 3.2475 x 10^-3 mol
Since both HNO3 and NaOH have a stoichiometry ratio of 1:1, the moles of nitric acid are also 3.2475 x 10^-3 mol.
Next, let's calculate the concentration of the nitric acid solution using the formula:
concentration (mol/L) = moles / volume (L)
= 3.2475 x 10^-3 mol / 0.02500 L
= 0.1133 mol/L
The concentration of the nitric acid solution is 0.1133 mol/L.