A titration is performed by adding 0.124 M KOH to 40 mL of 0.159 M HNO3.

a) Calculate the pH before addition of any KOH.













b) Calculate the pH after the addition of 10.26, 25.65 and 50.29 mL of the base.(Show your work in detail for one of the volumes.)

























c) Calculate the volume of base needed to reach the equivalence point.

HNO3 + KOH ==> KNO3 + H2O

a)HNO3 is 100% ionized; therefore, pH = -log(H^+) = -log(HNO3) = ?

b)
mols HNO3 = M x L = ?
mols KOH added = ?
mols HNO3-mols KOH = mols HNO3 remaining.
M HNO3 remaining = mols HNO3/total volume. Don't forget total volume will be amount HNO3 you started with + KOH volume added.

c.
mols HNO3 = M x L
mols KOH needed = mols HNO3 (see the coefficients in the balanced equation.)
mols KOH = M x L. You know mols and M, solve for L. Convert to mL if needed.

a) To calculate the pH before the addition of any KOH, we can use the equation for the neutralization reaction between HNO3 and KOH:

HNO3 + KOH → H2O + KNO3

Since the stoichiometric ratio between HNO3 and KOH is 1:1, we can determine the number of moles of HNO3 present in the solution:

mol HNO3 = Molarity x Volume
= 0.159 M x 0.040 L
= 0.00636 mol HNO3

Since the volume of KOH added is zero at the beginning, the number of moles of KOH present in the solution is also zero.

Now, we can calculate the concentration of HNO3 in the presence of KOH:

Molarity HNO3 = (moles HNO3) / (volume solution + volume KOH)
= 0.00636 mol / (0.040 L + 0 L)
= 0.00636 M

Using the concentration of HNO3, we can calculate the pH using the formula:

pH = -log10(Molarity)
= -log10(0.00636)
≈ 2.196

Therefore, the pH before the addition of any KOH is approximately 2.196.

b) To calculate the pH after the addition of a specific volume of base, we need to consider the reaction between HNO3 and KOH. Since this reaction is a strong acid-strong base neutralization reaction, the resulting solution will be a salt (KNO3) and water (H2O):

HNO3 + KOH → H2O + KNO3

We need to determine the moles of HNO3 and KOH used in the neutralization reaction.

Given that the initial volume of HNO3 is 40 mL and the molarity of HNO3 is 0.159 M:

moles HNO3 = Molarity x Volume
= 0.159 M x 0.040 L
= 0.00636 mol HNO3

Since KOH is added in different volumes, we need to calculate the moles of KOH used for each volume.

For example, let's calculate the moles of KOH used when 10.26 mL of base is added:

moles KOH = Molarity x Volume
= 0.124 M x 0.01026 L
= 0.00127344 mol KOH

Next, we need to determine the resulting moles of HNO3 and KOH after neutralization:

moles HNO3 remaining = initial moles HNO3 - moles KOH used
= 0.00636 mol - 0.00127344 mol
= 0.00508656 mol HNO3

moles KOH remaining = initial moles KOH - moles KOH used
= 0 mol KOH - 0.00127344 mol
= -0.00127344 mol KOH (negative value means KOH is in excess)

To determine the concentration of the remaining HNO3, we divide the moles by the total volume of the solution at each step:

Molarity HNO3 remaining = (moles HNO3 remaining) / (total volume of solution)
= 0.00508656 mol / (0.040 L + 0.01026 L)
≈ 0.111 M

Using this concentration, we can calculate the pH after the addition of 10.26 mL of KOH using the formula:

pH = -log10(Molarity)
= -log10(0.111)
≈ 0.954

Therefore, the pH after the addition of 10.26 mL of KOH is approximately 0.954.

c) To calculate the volume of base needed to reach the equivalence point, we need to determine the stoichiometry of the reaction between HNO3 and KOH. Since both are monoprotic acids and bases, the stoichiometry is 1:1.

Given that the molarity of HNO3 is 0.159 M and the volume of HNO3 is 40 mL:

moles HNO3 = Molarity x Volume
= 0.159 M x 0.040 L
= 0.00636 mol HNO3

Since the stoichiometric ratio is 1:1, the moles of KOH required will be equal to the moles of HNO3:

moles KOH = 0.00636 mol

To calculate the volume of KOH required, we divide the moles of KOH by its molarity:

volume KOH = moles KOH / Molarity
= 0.00636 mol / 0.124 M
≈ 0.051 L

Therefore, the volume of KOH needed to reach the equivalence point is approximately 0.051 L or 51 mL.