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Chapter 6: Problem 79

A student titrates an unknown amount of potassium hydrogen phthalate \(\left(\mathrm{KHC}_{8} \mathrm{H}_{4} \mathrm{O}_{4},\) often abbreviated KHP) with \right. \(20.46 \mathrm{mL}\) of a \(0.1000-M \mathrm{NaOH}\) solution. KHP (molar mass \(=\) \(204.22 \mathrm{g} / \mathrm{mol}\) ) has one acidic hydrogen. What mass of KHP was titrated (reacted completely) by the sodium hydroxide solution?

Short Answer

Expert verified
The mass of KHP titrated (reacted completely) by the sodium hydroxide solution is approximately \(0.418 \mathrm{g}\).

Step by step solution

01

Calculate moles of NaOH used in the titration

First, we will calculate the moles of NaOH used in the titration using the given volume and molarity. moles of NaOH = volume of NaOH (in L) × molarity of NaOH moles of NaOH = \(20.46 \times 10^{-3} \mathrm{L} \times 0.1000 \frac{\mathrm{mol}}{\mathrm{L}}\)
02

Determine moles of KHP that reacted

Since the reaction between KHP and NaOH is in a 1:1 ratio, the moles of KHP reacted will be equal to the moles of NaOH used in the titration. moles of KHP = moles of NaOH
03

Calculate the mass of KHP titrated

Now, using the molar mass of KHP, we can determine the mass of KHP titrated. mass of KHP = moles of KHP × molar mass of KHP mass of KHP = moles of KHP × \(204.22 \frac{\mathrm{g}}{\mathrm{mol}}\)
04

Calculate the final results

Plug the values from previous steps into the equations and solve for mass of KHP. moles of NaOH = \(20.46 \times 10^{-3} \mathrm{L} \times 0.1000 \frac{\mathrm{mol}}{\mathrm{L}} = 2.046 \times 10^{-3}\) mol moles of KHP = moles of NaOH = \(2.046 \times 10^{-3}\) mol mass of KHP = moles of KHP × \(204.22 \frac{\mathrm{g}}{\mathrm{mol}} = 2.046 \times 10^{-3} \mathrm{mol} \times 204.22 \frac{\mathrm{g}}{\mathrm{mol}} \approx 0.418 \mathrm{g}\) The mass of KHP titrated (reacted completely) by the sodium hydroxide solution is approximately 0.418 g.

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Most popular questions from this chapter

In the spectroscopic analysis of many substances, a series of standard solutions of known concentration are measured to generate a calibration curve. How would you prepare standard solutions containing \(10.0,25.0,50.0,75.0,\) and \(100 .\) ppm of copper from a commercially produced 1000.0 -ppm solution? Assume each solution has a final volume of \(100.0 \mathrm{mL}\). (See Exercise 123 for definitions.)

A sample may contain any or all of the following ions: \(\mathrm{Hg}_{2}^{2+}\) \(\mathrm{Ba}^{2+},\) and \(\mathrm{Mn}^{2+}\) a. No precipitate formed when an aqueous solution of \(\mathrm{NaCl}\) was added to the sample solution. b. No precipitate formed when an aqueous solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) was added to the sample solution. c. A precipitate formed when the sample solution was made basic with NaOH. Which ion or ions are present in the sample solution?

Write the balanced formula, complete ionic, and net ionic equations for each of the following acid-base reactions. a. \(\mathrm{HNO}_{3}(a q)+\mathrm{Al}(\mathrm{OH})_{3}(s) \rightarrow\) b. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{KOH}(a q) \rightarrow\) c. \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{HCl}(a q) \rightarrow\)

A 1.42 -g sample of a pure compound, with formula \(\mathrm{M}_{2} \mathrm{SO}_{4}\) was dissolved in water and treated with an excess of aqueous calcium chloride, resulting in the precipitation of all the sulfate ions as calcium sulfate. The precipitate was collected, dried, and found to weigh \(1.36 \mathrm{g}\). Determine the atomic mass of \(\mathrm{M},\) and identify \(\mathrm{M}\).

What mass of solid aluminum hydroxide can be produced when 50.0 mL of \(0.200 M\) Al(NO_j) is added to 200.0 mL of \(0.100 M\) KOH?
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