A solution containing nitric acid (HNO3) is to be used in the neutralization titration of a sodium hydroxide (NaOH) solution.
a. If the pH of the nitric acid solution is 2.30, what is the normality of the nitric acid?
b. If 27.00 ml of the nitric acid solution are required to neutralize 45.00 ml of the NaOH solution, what is the normality of the NaOH solution?
c. What was the pH of the NaOH solution before neutralization?
a. pH = -log(HNO3)
2.30 -log(HNO3)
(HNO3) = ? in moles/L which is the same as equivalents/L = normality.
b. mLNaOH x N NaOH = mL HNO3 x N HNO3
You know mL NaOH, mL HNO3, and N HNO3. Solve for N NaOH
c. pOH = -log(NaOH)
Post your work if you get stuck.
Then pH + pOH = pKw = 14
You know pKw and you know pOH, solve for pH.
a. To find the normality (N) of the nitric acid solution, we need to use the formula:
Normality (N) = (Molarity (M)) x (number of hydrogen ions (H+))
In the case of nitric acid (HNO3), there is only one hydrogen ion, so the number of hydrogen ions is 1.
Given that the pH of the nitric acid solution is 2.30, we can convert it to molarity using the formula:
[H+] = 10^(-pH)
[H+] = 10^(-2.30) ≈ 0.0045 M
Now we can calculate the normality:
Normality (N) = (0.0045 M) x 1 = 0.0045 N
Therefore, the normality of the nitric acid solution is 0.0045 N.
b. The formula to calculate the normality (N) of the NaOH solution is the same as before:
Normality (N) = (Molarity (M)) x (number of hydroxide ions (OH-))
In the case of sodium hydroxide (NaOH), there is only one hydroxide ion, so the number of hydroxide ions is 1.
Given that 27.00 ml of the nitric acid solution neutralizes 45.00 ml of the NaOH solution, we can set up the following equation using the normality relationship:
N(HNO3) x V(HNO3) = N(NaOH) x V(NaOH)
Using the values given:
0.0045 N x 27.00 ml = N(NaOH) x 45.00 ml
Now we can solve for the normality of the NaOH solution:
N(NaOH) = (0.0045 N x 27.00 ml) / 45.00 ml
N(NaOH) ≈ 0.0027 N
Therefore, the normality of the NaOH solution is 0.0027 N.
c. The pH of the NaOH solution before neutralization can be determined using the pOH formula:
pOH = -log([OH-])
Since we know that NaOH is a strong base that completely dissociates in water, we can assume that the concentration of hydroxide ions is equal to the concentration of NaOH:
[OH-] = 0.0027 N
Taking the negative logarithm:
pOH = -log(0.0027)
pOH ≈ 2.57
Now we can convert the pOH to pH using the relationship:
pH + pOH = 14
pH = 14 - 2.57
pH ≈ 11.43
Therefore, the pH of the NaOH solution before neutralization is approximately 11.43.
a. To determine the normality of the nitric acid solution, we need to understand the concept of normality and how it relates to pH.
Normality (N) is a concentration unit that indicates the number of equivalents of a solute per liter of a solution. An equivalent is the amount of a substance that can react with or replace 1 mole of hydrogen ions (H+) in a chemical reaction.
To find the normality of the nitric acid solution, we can use the formula:
Normality (N) = (Molarity (M) × Number of H+ ions) / (# of equivalents of solute).
In nitric acid (HNO3), there is only one H+ ion that can react. The number of equivalents of nitric acid is equal to the number of moles of H+ ions.
To determine the molarity (M) of nitric acid from pH, we can use the fact that pH is related to the concentration of H+ ions in a solution. The pH scale is a logarithmic scale that ranges from 0 to 14, where pH 7 is considered neutral (equal concentrations of H+ and OH- ions). pH values below 7 indicate acidic solutions, and pH values above 7 indicate basic (alkaline) solutions.
The relationship between pH and the concentration of H+ ions can be expressed as:
[H+] = 10^(-pH).
Given that the pH of the nitric acid solution is 2.30, we can find the concentration of H+ ions:
[H+] = 10^(-2.30) = 0.004466 moles per liter (M).
Since the concentration of H+ ions is also the molarity of the nitric acid solution, the normality can be determined as:
Normality (N) = (0.004466 M × 1 H+ ion) / (1 equivalent of solute)
Therefore, the normality of the nitric acid solution is 0.004466N.
b. To determine the normality of the NaOH solution, we can use the concept of stoichiometry. In a neutralization reaction between an acid and a base, the ratio of moles of acid to moles of base is 1:1. Therefore, the number of moles of nitric acid used to neutralize the NaOH solution is equal to the number of moles of NaOH.
Given that 27.00 ml of the nitric acid solution neutralizes 45.00 ml of the NaOH solution, we can calculate the number of moles of nitric acid used:
Moles of HNO3 = Molarity (M) × Volume (L) = 0.004466 M × (27.00 ml / 1000) = 0.00012 moles.
Since the number of moles of acid and base is equal, the moles of NaOH can be calculated:
Moles of NaOH = 0.00012 moles.
The volume of the NaOH solution used is 45.00 ml, which is equal to 0.045 L. Therefore, the molarity (M) of the NaOH solution is:
Molarity (M) = Moles of NaOH / Volume (L) = 0.00012 moles / 0.045 L = 0.0027 M.
The molarity of NaOH is also the normality of NaOH since NaOH is a monoprotic base, meaning it donates only one hydroxide ion (OH-) per mole.
Therefore, the normality of the NaOH solution is 0.0027N.
c. To determine the pH of the NaOH solution before neutralization, we need to understand that NaOH is a strong base that fully dissociates into its ions in solution. So before neutralization, the concentration of OH- ions is equal to the concentration of NaOH.
Using the molarity of the NaOH solution calculated in part b (0.0027 M), the concentration of OH- ions is 0.0027 moles per liter (M).
We can calculate the pOH of the NaOH solution using the relationship between pOH and the concentration of OH- ions:
pOH = -log[OH-] = -log(0.0027) ≈ 2.57.
Since pH + pOH = 14 in aqueous solutions, we can find the pH:
pH = 14 - pOH = 14 - 2.57 ≈ 11.43.
Therefore, the pH of the NaOH solution before neutralization is approximately 11.43.