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A 56.3-mL sample of 0.225 M HNO3 is titrated with 0.156 M KOH.

a. Calculate the volume of base required (in mL) to reach the equivalence point.

b. Calculate the pH after adding 98.5 mL of KOH

Answer :

a. The volume of base required to reach the equivalence point is 81.2 mL.
b. After adding 98.5 mL of KOH, the pH of the solution will be 7.

a. To calculate the volume of base required to reach the equivalence point, we can use the equation:

moles of acid = moles of base

First, we need to calculate the moles of acid. We have a 56.3 mL sample of 0.225 M HNO3. To find the moles, we can use the formula:

moles = volume (in L) × concentration (in mol/L)

Converting the volume to liters:

56.3 mL = 56.3 mL × (1 L / 1000 mL) = 0.0563 L

Calculating the moles of acid:

moles of acid = 0.0563 L × 0.225 mol/L = 0.01267 mol

Since the balanced equation for the reaction is 1:1 between HNO3 and KOH, we know that the moles of base required will be the same as the moles of acid.

Now, let's calculate the volume of base required. We have a 0.156 M KOH solution. We can rearrange the formula to solve for volume:

volume (in L) = moles / concentration (in mol/L)

Substituting the values:

volume of base = 0.01267 mol / 0.156 mol/L = 0.0812 L

Finally, let's convert the volume to milliliters:

volume of base = 0.0812 L × (1000 mL / 1 L) = 81.2 mL

Therefore, the volume of base required to reach the equivalence point is 81.2 mL.

b. To calculate the pH after adding 98.5 mL of KOH, we need to determine the amount of excess KOH remaining in the solution.

First, let's calculate the moles of KOH used. We have a 0.156 M KOH solution and a volume of 98.5 mL. Converting the volume to liters:

98.5 mL = 98.5 mL × (1 L / 1000 mL) = 0.0985 L

Calculating the moles of KOH used:

moles of KOH = 0.0985 L × 0.156 mol/L = 0.01533 mol

Since the balanced equation is 1:1 between HNO3 and KOH, the moles of KOH used will be equal to the moles of HNO3 reacted.

Now, let's calculate the moles of HNO3 remaining. We initially had 0.01267 mol of HNO3, so the moles remaining will be:

moles of HNO3 remaining = initial moles - moles of KOH used
moles of HNO3 remaining = 0.01267 mol - 0.01533 mol = -0.00266 mol

Since we can't have negative moles, it means that all the HNO3 has reacted. Therefore, after adding 98.5 mL of KOH, there is no HNO3 remaining in the solution.

The pH of a solution depends on the concentration of H+ ions. Since there is no HNO3 remaining, the concentration of H+ ions is zero, resulting in a pH of 7 (neutral).

Therefore, after adding 98.5 mL of KOH, the pH of the solution will be 7.

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