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Chapter 14: Problem 14

A buffer solution is prepared by adding \(5.50 \mathrm{~g}\) of ammonium chloride and \(0.0188\) mol of ammonia to enough water to make \(155 \mathrm{~mL}\) of solution. (a) What is the \(\mathrm{pH}\) of the buffer? (b) If enough water is added to double the volume, what is the \(\mathrm{pH}\) of the solution?

Short Answer

Expert verified
Answer: (a) The pH of the buffer solution is 8.30. (b) The pH of the buffer solution after doubling the volume is 7.96.

Step by step solution

01

Find the concentrations of ammonium chloride and ammonia in the original solution.

To do this, we will find the moles of ammonium chloride and then divide it by the volume of the solution. Ammonium chloride: mass = 5.50 g Molar mass of NH4Cl = 53.49 g/mol moles of NH4Cl = mass / molar mass moles of NH4Cl = 5.50 g / 53.49 g/mol = 0.1028 mol Ammonia: moles of NH3 = 0.0188 mol Now, we will find the concentrations of each component in the solution. Volume of solution = 155 mL = 0.155 L Concentration of NH4Cl = moles of NH4Cl / volume of solution Concentration of NH4Cl = 0.1028 mol / 0.155 L = 0.663 M Concentration of NH3 = moles of NH3 / volume of solution Concentration of NH3 = 0.0188 mol / 0.155 L = 0.121 M
02

Use the Henderson-Hasselbalch equation to find the pH of the buffer solution.

The Henderson-Hasselbalch equation is: pH = pKa + log ([base]/[acid]) The reaction for this buffer solution is: NH3 + H2O <-> NH4+ + OH- pKa = -log(Ka) (for ammonia, NH3) The Ka value for ammonia is 1.8 × 10^(-5), so pKa = -log(1.8 × 10^(-5)) = 9.74 Now, we will find the pH: pH = 9.74 + log (0.121/0.663) pH = 9.74 - 1.44 = 8.30 The pH of the buffer solution is 8.30.
03

Find the new pH after diluting the buffer solution.

When we double the volume of the solution, the concentrations of NH4Cl and NH3 will be halved. New concentration of NH4Cl = 0.663 M / 2 = 0.3315 M New concentration of NH3 = 0.121 M / 2 = 0.0605 M Using the Henderson-Hasselbalch equation again: pH = 9.74 + log (0.0605/0.3315) pH = 9.74 - 1.78 = 7.96 The pH of the diluted buffer solution is 7.96. Answer: (a) The pH of the buffer solution is 8.30. (b) The pH of the buffer solution after doubling the volume is 7.96.

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Most popular questions from this chapter

Phenol red is an indicator with a \(\mathrm{pK}_{\mathrm{a}}\) of \(7.4\). It is yellow in acid solution and red in alkaline. (a) What is its \(K_{\mathrm{a}}\) ? (b) What is its \(\mathrm{pH}\) range? (c) What would its color be at \(\mathrm{pH}\) 7.4?

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.)

Which of the following would form a buffer if added to \(650.0 \mathrm{~mL}\) of \(0.40 M \mathrm{Sr}(\mathrm{OH})_{2} ?\) (a) \(1.00 \mathrm{~mol}\) of \(\mathrm{HF}\) (b) \(0.75 \mathrm{~mol}\) of \(\mathrm{HF}\) (c) \(0.30 \mathrm{~mol}\) of \(\mathrm{HF}\) (d) \(0.30 \mathrm{~mol}\) of \(\mathrm{NaP}\) (e) \(0.30 \mathrm{~mol}\) of \(\mathrm{HCl}\) Explain your reasoning in each case.

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.)

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?
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