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Chapter 14: Problem 44

WEB A 25.00-mL sample of formic acid, \(\mathrm{HCHO}_{2}\), is titrated with \(39.74 \mathrm{~mL}\) of \(0.117 \mathrm{M} \mathrm{KOH}\). (a) What is \(\left[\mathrm{HCHO}_{2}\right]\) before titration? (b) Calculate \(\left[\mathrm{HCHO}_{2}\right],\left[\mathrm{CHO}_{2}^{-}\right],\left[\mathrm{OH}^{-}\right]\), and \(\left[\mathrm{K}^{+}\right]\) at the equivalence point. (Assume that volumes are additive.) (c) What is the \(\mathrm{pH}\) at the equivalence point?

Short Answer

Expert verified
Based on the given problem and step-by-step solution, provide a short answer regarding the concentration of formic acid before titration, the concentrations of different species at the equivalence point, and the pH at the equivalence point. The initial concentration of formic acid (HCHO₂) before titration is 0.186 M. At the equivalence point, the concentrations of potassium ions (K⁺), hydroxide ions (OH⁻), and formate ions (CHO₂⁻) are all 0.0718 M. The pH at the equivalence point is 12.86.

Step by step solution

01

Calculate the moles of KOH used in the titration

To find the moles of KOH, we can use the given volume and molarity:$$ \text{moles of KOH} = \text{volume} \times \text{molarity} = 39.74\,\text{mL} \times 0.117\,\frac{\text{moles}}{\text{L}}$$We need to convert mL to L:$$ 39.74\,\text{mL} \times \frac{1\,\text{L}}{1000\,\text{mL}} = 0.03974\,\text{L}$$Now, we multiply the volume by the molarity:$$ \text{moles of KOH} = 0.03974\,\text{L} \times 0.117\,\frac{\text{moles}}{\text{L}} = 0.00465\,\text{moles} $$
02

Determine the initial concentration of HCHO₂

At the equivalence point, the moles of HCHO₂ and KOH are equal. So, we can write:$$ \text{moles of HCHO₂} = \text{moles of KOH} = 0.00465\,\text{moles} $$To find the initial concentration of HCHO₂, divide the moles by the initial volume:$$ \left[\mathrm{HCHO}_{2}\right] = \frac{\text{moles of HCHO₂}}{\text{initial volume}} = \frac{0.00465\,\text{moles}}{0.025\,\text{L}} = 0.186\,\frac{\text{moles}}{\text{L}} $$
03

Calculate the concentrations of all species at the equivalence point

At the equivalence point, all the HCHO₂ has reacted with the KOH to form CHO₂⁻ and water. So, we have:$$ \left[\mathrm{HCHO}_{2}\right] = 0 \\ \left[\mathrm{K}^{+}\right] = \left[\mathrm{OH}^{-}\right] = \left[\mathrm{CHO}_{2}^{-}\right] \\ $$To find the concentrations of these ions, we can use the moles of KOH and the total volume at the equivalence point (initial volume of HCHO₂ + volume of KOH):$$ \text{total volume} = 0.025\,\text{L} + 0.03974\,\text{L} = 0.06474\,\text{L} $$Now, we can find the concentrations:$$ \left[\mathrm{K}^{+}\right] = \left[\mathrm{OH}^{-}\right] = \left[\mathrm{CHO}_{2}^{-}\right] = \frac{0.00465\,\text{moles}}{0.06474\,\text{L}} = 0.0718\,\frac{\text{moles}}{\text{L}} $$
04

Find the pH at the equivalence point

At the equivalence point, we know [OH⁻] and can use it to find the pOH. Then, we can find the pH using the relationship between pH and pOH:$$ \text{pOH} = -\log\left[\mathrm{OH}^{-}\right] = -\log(0.0718) \approx 1.14 \\ \text{pH} = 14 - \text{pOH} = 14 - 1.14 = 12.86 $$So, the pH at the equivalence point is 12.86.

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Most popular questions from this chapter

A solution of \(\mathrm{NaOH}\) with \(\mathrm{pH} 13.68\) requires \(35.00 \mathrm{~mL}\) of \(0.128 \mathrm{M}\) \(\mathrm{HClO}_{4}\) to reach the equivalence point. (a) What is the volume of the \(\mathrm{NaOH}\) solution? (b) What is the \(\mathrm{pH}\) at the equivalence point? that volumes are additive.)

There is a buffer system \(\left(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}-\mathrm{HPO}_{4}{ }^{2-}\right)\) in blood that helps keep the blood \(\mathrm{pH}\) at about \(7.40 .\left(\mathrm{K}_{\mathrm{a}} \mathrm{H}_{2} \mathrm{PO}_{4}^{-}=6.2 \times 10^{-8}\right)\). (a) Calculate the \(\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right] /\left[\mathrm{HPO}_{4}^{2-}\right]\) ratio at the normal \(\mathrm{pH}\) of blood. (b) What percentage of the \(\mathrm{HPO}_{4}{ }^{2-}\) ions are converted to \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) when the \(\mathrm{pH}\) goes down to \(6.80\) ? (c) What percentage of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ions are converted to \(\mathrm{HPO}_{4}{ }^{2-}\) when the \(\mathrm{pH}\) goes up to \(7.80\) ?

Consider the titration of \(\mathrm{HF}\left(K_{\mathrm{a}}=6.7 \times 10^{-4}\right)\) with \(\mathrm{NaOH}\). What is the \(\mathrm{pH}\) when a third of the acid has been neutralized?

WEB To make a buffer with \(\mathrm{pH}=3.0\) from \(\mathrm{HCHO}_{2}\) and \(\mathrm{CHO}_{2}^{-}\), (a) what must the [HCHO \(\left._{2}\right] /\left[\mathrm{CHO}_{2}^{-}\right]\) ratio be? (b) how many moles of \(\mathrm{HCHO}_{2}\) must be added to a liter of \(0.139 \mathrm{M}\) \(\mathrm{NaCHO}_{2}\) to give this \(\mathrm{pH}\) ? (c) how many grams of \(\mathrm{NaCHO}_{2}\) must be added to \(350.0 \mathrm{~mL}\) of \(0.159 \mathrm{MHCHO}_{2}\) to give this \(\mathrm{pH}\) ? (d) What volume of \(0.236 \mathrm{M} \mathrm{HCHO}_{2}\) must be added to \(1.00 \mathrm{~L}\) of a \(0.500 \mathrm{M}\) solution of \(\mathrm{NaCHO}_{2}\) to give this pH? (Assume that volumes are additive.)

Four grams of a monoprotic weak acid are dissolved in water to make \(250.0 \mathrm{~mL}\) of solution with a pH of \(2.56\). The solution is divided into two equal parts, \(\mathrm{A}\) and \(\mathrm{B}\). Solution \(\mathrm{A}\) is titrated with strong base to its equivalence point. Solution B is added to solution A after solution \(\mathrm{A}\) is neutralized. The \(\mathrm{pH}\) of the resulting solution is \(4.26\). What is the molar mass of the acid?
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