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!!!:)
(d) Determine the pH of a partially dissolved solution of phosphorous acid at a concentration of 0.0025 mol/L and a volume of 50.0 mL.
(e) What is the concentration of OH ions in this solution?
1) To find the volume of hydrobromic acid needed to neutralize the magnesium hydroxide solution, we need to use the balanced chemical equation:
2HBr(aq) + Mg(OH)2(aq) -> MgBr2(aq) + 2H2O(l)
From the equation, we can see that the stoichiometric ratio between HBr and Mg(OH)2 is 2:1. This means that for every 2 moles of HBr, we react with 1 mole of Mg(OH)2.
First, let's calculate the moles of Mg(OH)2 present in 75.0 mL of the solution:
Molarity of Mg(OH)2 = 0.895 mol/L
Volume of Mg(OH)2 = 75.0 mL = 0.075 L
Moles of Mg(OH)2 = Molarity * Volume = 0.895 mol/L * 0.075 L = 0.0671 moles
Since the stoichiometric ratio between HBr and Mg(OH)2 is 2:1, we need half the number of moles of HBr to neutralize the Mg(OH)2.
Moles of HBr needed = 0.0671 moles / 2 = 0.0336 moles
Now, we can calculate the volume of the HBr solution using its molarity:
Molarity of HBr = 1.25 mol/L
Volume of HBr = Moles / Molarity = 0.0336 moles / 1.25 mol/L = 0.0269 L or 26.9 mL
Therefore, the volume of the 1.25 mol/L hydrobromic acid solution needed to neutralize the 75.0 mL of the 0.895 mol/L magnesium hydroxide solution is approximately 26.9 mL.
2) To determine the concentration of the nitric acid solution, we can use the equation:
HNO3(aq) + NaOH(aq) -> NaNO3(aq) + H2O(l)
From the equation, we can see that the stoichiometric ratio between HNO3 and NaOH is 1:1. This means that for every 1 mole of HNO3, we react with 1 mole of NaOH.
Let's first calculate the moles of NaOH used in the neutralization reaction:
Molarity of NaOH = 0.1750 mol/L
Volume of NaOH = 18.55 mL = 0.01855 L
Moles of NaOH = Molarity * Volume = 0.1750 mol/L * 0.01855 L = 0.00324 moles
Since the stoichiometric ratio between HNO3 and NaOH is 1:1, the moles of HNO3 used in the reaction is also 0.00324 moles.
Now, we can calculate the concentration of the nitric acid solution using its volume:
Volume of HNO3 = 25.00 mL = 0.02500 L
Molarity of HNO3 = Moles / Volume = 0.00324 moles / 0.02500 L = 0.1296 mol/L or 0.130 mol/L (rounded to three significant figures)
Therefore, the concentration of the nitric acid solution is approximately 0.130 mol/L.
Sure! I'd be happy to help you with these two questions.
1) To find the volume of the hydrobromic acid, HBr(aq), that will neutralize the given volume of magnesium hydroxide, Mg(OH)2(aq), you can use the equation:
Acid volume × Acid concentration = Base volume × Base concentration
Let's denote the volume of hydrobromic acid as V(HBr) and the volume of magnesium hydroxide as V(Mg(OH)2).
According to the equation above, we have:
V(HBr) × 1.25 mol/L = 75.0 mL × 0.895 mol/L
We can rearrange this equation to solve for V(HBr):
V(HBr) = (75.0 mL × 0.895 mol/L) / 1.25 mol/L
V(HBr) ≈ 53.76 mL
Therefore, the volume of the 1.25 mol/L hydrobromic acid needed to neutralize 75.0 mL of 0.895 mol/L magnesium hydroxide is approximately 53.76 mL.
2) To find the concentration of the nitric acid solution, HNO3(aq), you can use the equation:
Acid volume × Acid concentration = Base volume × Base concentration
Let's denote the concentration of the nitric acid as C(HNO3).
According to the given information, we have:
25.00 mL × C(HNO3) = 18.55 mL × 0.1750 mol/L
We can rearrange this equation to solve for C(HNO3):
C(HNO3) = (18.55 mL × 0.1750 mol/L) / 25.00 mL
C(HNO3) ≈ 0.1298 mol/L
Therefore, the concentration of the nitric acid solution is approximately 0.1298 mol/L.
I hope this helps you with your questions! Let me know if there's anything else I can assist you with.
2HBr + Mg(OH)2 ==> MgBr2 + 2H2O
mols Mg(OH)2 = M x L = ?
Convert mols Mg(OH) to mols HBr using the coefficients.
Then M HBr = mols HBr/L HBr. You know mols and M, solve for L.
The second problem is work the same as the first