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Chapter 16: Problem 134

A \(25.0-\mathrm{mL}\) of \(0.20 \mathrm{M}\) HF solution is titrated with a \(0.20 M\) NaOH solution. Calculate the volume of \(\mathrm{NaOH}\) solution added when the \(\mathrm{pH}\) of the solution is (a) \(2.85,\) (b) \(3.15,\) (c) \(11.89 .\) Ignore salt hydrolysis.

Short Answer

Expert verified
The volumes of \(0.20 M NaOH\) solution required to reach pH values of 2.85, 3.15 and 11.89 while titrating a \(0.20 M HF\) solution are approximately 5.0 mL, 12.5 mL and more than the initial volume of HF solution respectively. However, it is impossible to reach a pH of 11.89 without considering salt hydrolysis.

Step by step solution

01

Find the \(pKa\)

Because HF is a weak acid, it will have a \(pKa\) value (the negative log of the acid dissociation constant \(Ka\)). For HF, the \(pKa\) is approximately 3.17.
02

Determine amount of HF in moles

This can be determined using the formula \(Molarity (M) = Moles/Liters\) or in other words Moles = \(Molarity * Volume\). In this case, Moles of HF = \(0.20 M * 0.025 L = 0.005 moles\).
03

Calculate the volume of NaOH solution required to reach \(pH = 2.85\)

We use the Henderson-Hasselbalch equation \(pH = pKa + log ([A-]/[HA])\). Rearranging for [HA], we get [HA] = [A-] * \(10^(pKa - pH)\). As pH < pKa, [HA] would be more than [A-]. This means moles of HF that have not donated a proton are more than moles of NaOH added at this point. From the above, we get moles of NaOH added = moles of HF initially - moles of HF left = 0.005 - (0.005 * \(10^(3.17 - 2.85)\)) = approximately 0.001 moles. Therefore, the volume of 0.20 M NaOH solution required to reach this pH = moles/molarity = 0.001/0.20 = 0.005 L or 5.0 mL.
04

Calculate volume of NaOH solution required to reach \(pH = 3.15 (pKa)\)

At the point where pH = pKa, the moles of HF (HA) are equal to the moles of F- (A-), which means that half the moles of HF have been neutralised by the NaOH. Therefore, the volume of 0.20 M NaOH solution required to reach this pH = moles/molarity = 0.0025/0.20 = 0.0125 L or 12.5 mL.
05

Calculate volume of NaOH solution required to reach \(pH = 11.89\)

Now that the pH is greater than the \(pKa\), we know we have surpassed the equivalence point. Subtract moles of HF from the moles of NaOH added after the equivalence point. This gives the concentration of OH- left. From this, calculate pOH = -log[OH-], and subtract from 14 to get pH. However, as we are ignoring salt hydrolysis in this problem, this scenario isn't possible.

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