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Chapter 17: Problem 98

How many milliliters of \(1.0 \mathrm{M} \mathrm{NaOH}\) must be added to \(200 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{NaH}_{2} \mathrm{PO}_{4}\) to make a buffer solution with a pH of \(7.50 ?\)

Short Answer

Expert verified
19 ml of 1.0 M NaOH should be added to the solution to achieve a pH of 7.50.

Step by step solution

01

Understand the equation to use

The pH of a buffer solution can be calculated by the Henderson-Hasselbalch equation: pH = pKa + log ([A-]/[HA]). Here, [A-] is the concentration of conjugate base (in mol/L) and [HA] is the concentration of weak acid (in mol/L). We are given that the pH wanted is 7.50, and the pKa of \( H_2PO_4^- \) is 7.21 (from the table of Ka Values at 25 Degrees Celsius, given that the second pKa of phosphoric acid is 7.21).
02

Identify the concentrations provided

The initial concentration of \(NaH_2PO_4\) is given as 0.10 M in 200 mL. Therefore, the number of moles of \(NaH_2PO_4\) is given by: moles = concentration x volume = 0.10 mol/L x 0.2 L = 0.02 mol. Also, the concentration of \(NaOH\) (which will become A-) is given as 1.0 M.
03

Substitute in the Henderson-Hasselbalch Equation

We know pH = pKa + log ([A-]/[HA]), substituting the given and calculated values we get 7.50 = 7.21 + log([A-]/0.02). Solving this for [A-] we get [A-] = 0.019 mol/L. We can multiply this number by the volume of this solution in liters to obtain the number of moles of \(NaOH\) (or [A-]) needed. The volume is not given, so let's call it x liters.
04

Solve for the unknown

Calculate the volume of 1M \(NaOH\) solution needed to provide the required moles of \(NaOH\). Since the concentration of \(NaOH\) is 1M, the volume in liters is the same as the number of moles. Therefore, x= moles = 0.019 L = 19 mL. This means 19 mL of 1 M \(NaOH\) should be added to the solution to achieve a pH of 7.50.

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Most popular questions from this chapter

The \(\mathrm{p} K_{\mathrm{a}}\) of butyric acid (HBut) is 4.7 . Calculate \(K_{\mathrm{b}}\) for the butyrate ion (But \(^{-}\) ).

Calculate the \(\mathrm{pH}\) at the equivalence point for these titrations: (a) \(0.10 M \mathrm{HCl}\) versus \(0.10 \mathrm{M} \mathrm{NH}_{3}\), (b) \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) versus \(0.10 \mathrm{M} \mathrm{NaOH}\).

One way to distinguish a buffer solution with an acid solution is by dilution. (a) Consider a buffer solution made of \(0.500 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) and \(0.500 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COONa}\). Calculate its \(\mathrm{pH}\) and the \(\mathrm{pH}\) after it has been diluted 10 -fold. (b) Compare the result in (a) with the pHs of a \(0.500 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) solution before and after it has been diluted 10 -fold

\(\mathrm{CaSO}_{4}\left(K_{\mathrm{sp}}=2.4 \times 10^{-5}\right)\) has a larger \(K_{\mathrm{sp}}\) value than that of \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\left(K_{\mathrm{sp}}=1.4 \times 10^{-5}\right)\). Does it fol- low that \(\mathrm{CaSO}_{4}\) also has greater solubility \((\mathrm{g} / \mathrm{L}) ?\)

How can we predict whether a precipitate will form when two solutions are mixed?
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