help...If exactly 5.0 mL of HNO3 will neutralize 15 mL of 2.0 M NaOH, what is the molarity of the HNO3 solution?
Have no idea what to do.
thank you.
HNO3 + NaOH ==> NaNO3 + H2O
mols NaOH = M x L = ?
Look at the equation. It is 1:1 for base/acid so
mols HNO3 = mols NaOH
The M HNO3 = moles HNO3/L HNO3.
To solve this problem, you can use the concept of stoichiometry. The balanced chemical equation for the reaction of HNO3 and NaOH is:
HNO3 + NaOH → NaNO3 + H2O
From the equation, you can see that 1 mole of HNO3 reacts with 1 mole of NaOH.
Given that 15 mL of 2.0 M NaOH is used, you can calculate the number of moles of NaOH used by using the formula:
moles of NaOH = volume (L) × concentration (M)
moles of NaOH = 0.015 L × 2.0 M
moles of NaOH = 0.030 moles
Since the reaction is 1:1, the number of moles of HNO3 used is also 0.030 moles.
Now, you can calculate the molarity of the HNO3 solution using the formula:
molarity (M) = moles/volume (L)
Given that the volume of HNO3 used is 5.0 mL, you need to convert it to liters:
volume (L) = 5.0 mL ÷ 1000 mL/L
volume (L) = 0.005 L
Now, you can calculate the molarity:
molarity (M) = 0.030 moles ÷ 0.005 L
molarity (M) = 6.0 M
Therefore, the molarity of the HNO3 solution is 6.0 M.
To find the molarity (M) of the HNO3 solution, you can use the concept of stoichiometry and the equation for the neutralization reaction between HNO3 and NaOH. The balanced equation for this reaction is:
HNO3 + NaOH → NaNO3 + H2O
From the equation, we can determine the stoichiometric ratio between HNO3 and NaOH is 1:1. This means that 1 mole of HNO3 reacts with 1 mole of NaOH.
Given that 15 mL of 2.0 M NaOH is used for the neutralization, we can determine the number of moles of NaOH:
Moles of NaOH = Molarity × Volume
= 2.0 M × 0.015 L
= 0.030 moles
From the stoichiometric ratio, we know that the number of moles of HNO3 is also 0.030 moles since the ratio is 1:1.
The volume of the HNO3 solution used is 5.0 mL, which is 0.005 L. Then, we can calculate the molarity of the HNO3 solution as follows:
Molarity of HNO3 = Moles of HNO3 / Volume of HNO3 solution
= 0.030 moles / 0.005 L
= 6.0 M
Therefore, the molarity of the HNO3 solution is 6.0 M.