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Chapter 17: Problem 39

Calculate the \(\mathrm{pH}\) at the points in the titration of \(25.00 \mathrm{mL}\) of 0.160 M HCl when (a) 10.00 mL and (b) 15.00 mL of 0.242 M KOH have been added.

Short Answer

Expert verified
The pH at the points in the titration of 25.00 mL of 0.160 M HCl when (a) 10.00 mL and (b) 15.00 mL of 0.242 M KOH have been added are (a) 1.346 and (b) 11.966.

Step by step solution

01

Determine the mole of HCl and KOH

Calculate the moles of HCl and KOH using the molarity and volume given. The moles of acid is \(0.160 \, M \times 25.00 \, mL = 4.00 \times 10^{-3} \, moles\) and the moles of base for part (a) is \(0.242 \, M \times 10.00 \, mL = 2.42 \times 10^{-3} \, moles\) and for part (b) \(0.242 \, M \times 15.00 \, mL = 3.63 \times 10^{-3} \, moles\).
02

Calculate moles of remaining substance in part (a)

Subtract the moles of base from the moles of acid to get \(4.00 \times 10^{-3} \, moles - 2.42 \times 10^{-3} \, moles = 1.58 \times 10^{-3} \, moles\) of HCl left.
03

Calculate pH in part (a)

This remaining HCl makes the solution acidic. The concentration is found by dividing the remaining moles by the total volume of solution in liters which is \(35.00 \, mL = 0.035 \, L\) to get \(0.0451 \, M\). The pH is found using the formula \(-log[H^{+}]\) which gives \(pH = 1.346\).
04

Calculate moles of remaining substance in part (b)

Subtract the moles of acid from the moles of base to get \(3.63 \times 10^{-3} \, moles - 4.00 \times 10^{-3} \, moles = -0.37 \times 10^{-3} \, moles\). The negative sign indicates that all the acid has been neutralized and there is excess base.
05

Calculate pOH and then pH in part (b)

The concentration of the excess OH- is found by dividing the absolute moles of excess base by the total volume of solution \(0.040 \, L\) which gives \(0.00925 \, M\). The pOH is calculated by \(-log[OH^{-}]\) which gives \(pOH = 2.034\). The pH is found using the formula \(pH = 14 - pOH\) which gives \(pH = 11.966\).

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Most popular questions from this chapter

The \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}-\mathrm{HPO}_{4}^{2-}\) combination plays a role in maintaining the pH of blood. (a) Write equations to show how a solution containing these ions functions as a buffer. (b) Verify that this buffer is most effective at \(\mathrm{pH} 7.2\) (c) Calculate the \(\mathrm{pH}\) of a buffer solution in which \(\left[\mathrm{H}_{2} \mathrm{PO}_{4}\right]=0.050 \mathrm{M}\) and \(\left[\mathrm{HPO}_{4}^{2-}\right]=0.150 \mathrm{M} .[\)Hint: Focus on the second step of the phosphoric acid ionization.]

A solution is prepared that is \(0.150 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) and \(0.250 \mathrm{M} \mathrm{NaHCOO}\) (a) Show that this is a buffer solution. (b) Calculate the pH of this buffer solution. (c) What is the final pH if 1.00 L of 0.100 M HCl is added to \(1.00 \mathrm{L}\) of this buffer solution?

Sodium phosphate, \(\mathrm{Na}_{3} \mathrm{PO}_{4},\) is made commercially by first neutralizing phosphoric acid with sodium carbonate to obtain \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\). The \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) is further neutralized to \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) with \(\mathrm{NaOH}\) (a) Write net ionic equations for these reactions. (b) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is a much cheaper base than is \(\mathrm{NaOH}\) Why do you suppose that \(\mathrm{NaOH}\) must be used as well as \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) to produce \(\mathrm{Na}_{3} \mathrm{PO}_{4} ?\)

Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2},\) is a somewhat stronger acid than water. Its ionization is represented by the equation \(\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HO}_{2}^{-}\) In \(1912,\) the following experiments were performed to obtain an approximate value of \(\mathrm{p} K_{\mathrm{a}}\) for this ionization at \(0^{\circ} \mathrm{C} .\) A sample of \(\mathrm{H}_{2} \mathrm{O}_{2}\) was shaken together with a mixture of water and 1 -pentanol. The mixture settled into two layers. At equilibrium, the hydrogen peroxide had distributed itself between the two layers such that the water layer contained 6.78 times as much \(\mathrm{H}_{2} \mathrm{O}_{2}\) as the 1 -pentanol layer. In a second experiment, a sample of \(\mathrm{H}_{2} \mathrm{O}_{2}\) was shaken together with 0.250 M NaOH(aq) and 1-pentanol. At equilibrium, the concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) was \(0.00357 \mathrm{M}\) in the 1-pentanol layer and 0.259 M in the aqueous layer. In a third experiment, a sample of \(\mathrm{H}_{2} \mathrm{O}_{2}\) was brought to equilibrium with a mixture of 1 -pentanol and \(0.125 \mathrm{M}\) \(\mathrm{NaOH}(\mathrm{aq}) ;\) the concentrations of the hydrogen peroxide were \(0.00198 \mathrm{M}\) in the 1 -pentanol and \(0.123 \mathrm{M}\) in the aqueous layer. For water at \(0^{\circ} \mathrm{C}, \mathrm{p} K_{\mathrm{w}}=14.94\) Find an approximate value of \(\mathrm{pK}_{\mathrm{a}}\) for \(\mathrm{H}_{2} \mathrm{O}_{2}\) at \(0^{\circ} \mathrm{C}\) [Hint: The hydrogen peroxide concentration in the aqueous layers is the total concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) and \(\mathrm{HO}_{2}^{-}\). Assume that the 1 -pentanol solutions contain no ionic species.

Indicate whether you would expect the equivalence point of each of the following titrations to be below, above, or at \(\mathrm{pH}\) 7. Explain your reasoning. (a) \(\mathrm{NaHCO}_{3}(\mathrm{aq})\) is titrated with \(\mathrm{NaOH}(\mathrm{aq})\) (a) (b) \(\mathrm{HCl}(\mathrm{aq})\) is titrated with \(\mathrm{NH}_{3}(\mathrm{aq}) ;\) (c) \(\mathrm{KOH}(\mathrm{aq})\) is titrated with HI(aq).
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