My head is spinning from this chapter! I've done most of the questions by myself but I have no idea what formulas I should use for these last two. Can you walk me through them so I can understand what I need to do?
1. What volume of 5.0 M HNO3 is required to neutralize 2500 cm3 of a 2.0 M NaOH solution?
2. Calculate the volume of 0.55M HNO3 necessary to neutralize 5500 cm3 of 0.45M KOH.
1. Write and balance the equation.
HNO3 + NaOH ==> NaNO3 + H2O
2. mols NaOH = M x L = 2.0 M x 2.500 L = 5.0
3. Using the coefficients in the balanced equation convert mols NaOH to mols HNO3.
5.0 mols NaOH x (1 mol HNO3/1 mol NaOH) = 5.0 x 1/1 = 5.0 mols HNO3
4. M HNO3 = mols HNO3/L HNO3. You know M and mols, solve for L (convert to mL if needed).
The second problem is the same kind.
Sure! I'd be happy to help you understand how to approach these problems.
1. To solve this problem, you will need to use the concept of stoichiometry and the equation for the reaction between HNO3 and NaOH. The balanced chemical equation for the reaction is:
HNO3 + NaOH -> NaNO3 + H2O
First, let's identify the balanced equation for this neutralization reaction between HNO3 and NaOH. According to the equation, 1 mole of HNO3 reacts with 1 mole of NaOH, producing 1 mole of NaNO3 and 1 mole of water.
Now, let's determine the number of moles of NaOH in the 2500 cm3 (or 2.5 L) of the 2.0 M NaOH solution. To do this, use the formula:
moles = concentration (M) x volume (L)
moles = 2.0 M x 2.5 L = 5.0 moles of NaOH
Since the ratio between HNO3 and NaOH in the balanced equation is 1:1, you will need 5.0 moles of HNO3 to react with the 5.0 moles of NaOH.
Next, we need to calculate the volume of 5.0 M HNO3 (unknown volume) that contains 5.0 moles of HNO3. Rearranging the formula for moles, we have:
volume = moles / concentration
volume = 5.0 moles / 5.0 M = 1.0 L or 1000 cm3
So, you would need 1000 cm3 (or 1.0 L) of 5.0 M HNO3 to neutralize 2500 cm3 of a 2.0 M NaOH solution.
2. Similar to the previous problem, we will apply the concept of stoichiometry and the balanced chemical equation for the reaction between HNO3 and KOH:
HNO3 + KOH -> KNO3 + H2O
First, calculate the number of moles of KOH in the 5500 cm3 (or 5.5 L) of the 0.45 M KOH solution:
moles = concentration (M) x volume (L)
moles = 0.45 M x 5.5 L = 2.475 moles of KOH
Since the ratio between HNO3 and KOH in the balanced equation is 1:1, you will need 2.475 moles of HNO3 to react with the 2.475 moles of KOH.
Next, calculate the volume of the 0.55 M HNO3 (unknown volume) that contains 2.475 moles of HNO3:
volume = moles / concentration
volume = 2.475 moles / 0.55 M ≈ 4.50 L or 4500 cm3
Therefore, you would need approximately 4500 cm3 (or 4.50 L) of 0.55 M HNO3 to neutralize 5500 cm3 of 0.45 M KOH.
I hope this explanation helps you understand how to solve these types of problems! Let me know if you have any further questions.