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Chapter 4: Problem 87

A student carried out two titrations using a \(\mathrm{NaOH}\) solution of unknown concentration in the buret. In one titration she weighed out \(0.2458 \mathrm{~g}\) of KHP (see Section 4.7 ) and transferred it to an Erlenmeyer flask. She then added \(20.00 \mathrm{~mL}\) of distilled water to dissolve the acid. In the other titration she weighed out \(0.2507 \mathrm{~g}\) of KHP but added \(40.00 \mathrm{~mL}\) of distilled water to dissolve the acid. Assuming no experimental error, would she obtain the same result for the concentration of the \(\mathrm{NaOH}\) solution?

Short Answer

Expert verified
Yes, the student will obtain the same result for the concentration of the NaOH solution in both titrations, assuming no experimental errors and the same volume of NaOH was used for titration in both cases.

Step by step solution

01

Determine the moles of KHP

The molecular weight of KHP is \(204.22 g/mol\). The amount of moles of KHP in the first titration is \( \frac{0.2458 g }{ 204.22 g/mol } = 0.0012 mol \) and in the second titration is \( \frac{0.2507 g}{ 204.22 g/mol } = 0.00123 mol \)
02

Calculate the concentration of NaOH

The concentration of NaOH is equal to \( \frac{ moles of KHP }{Volume of NaOH solution used } \). Since the molar ratio between KHP and NaOH is 1:1, the same volume of NaOH was used in both titrations to completely neutralize the KHP. As these values of moles of KHP are very close to each other, the concentration of NaOH will be the same in both titrations if same volume of NaOH is used.
03

Number of moles of NaOH required for complete reaction

To validate the above result, we can calculate the number of moles of NaOH required for complete reaction with KHP. Using the stoichiometry of the reaction, 1 mole of KHP reacts with 1 mole of NaOH. Thus, 0.0012 mol of KHP will react with 0.0012 mol of NaOH in the first titration and 0.00123 mol of KHP will react with 0.00123 mol of NaOH in the second titration, given that the volume of NaOH used in both the titration is the same. The requirement of same amount of NaOH for complete reaction validates our earlier calculation of the same concentration of NaOH in both titrations.

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Most popular questions from this chapter

A sample of \(0.6760 \mathrm{~g}\) of an unknown compound containing barium ions \(\left(\mathrm{Ba}^{2+}\right)\) is dissolved in water and treated with an excess of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\). If the mass of the \(\mathrm{BaSO}_{4}\) precipitate formed is \(0.4105 \mathrm{~g},\) what is the percent by mass of Ba in the original unknown compound?

Which of the following aqueous solutions would you expect to be the best conductor of electricity at \(25^{\circ} \mathrm{C} ?\) Explain your answer. (a) \(0.20 \mathrm{M} \mathrm{NaCl}\) (b) \(0.60 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) (c) \(0.25 M \mathrm{HCl}\) (d) \(0.20 M \operatorname{Mg}\left(\mathrm{NO}_{3}\right)_{2}\)

Describe in each case how you would separate the cations or anions in an aqueous solution of: (a) \(\mathrm{NaNO}_{3}\) and \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2},\) (b) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{KNO}_{3}\), (c) \(\mathrm{KBr}\) and \(\mathrm{KNO}_{3}\), (d) \(\mathrm{K}_{3} \mathrm{PO}_{4}\) and \(\mathrm{KNO}_{3}\), (e) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{NaNO}_{3}\).

A \(46.2-\mathrm{mL}, 0.568 M\) calcium nitrate \(\left[\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\right]\) solution is mixed with \(80.5 \mathrm{~mL}\) of \(1.396 \mathrm{M}\) calcium nitrate solution. Calculate the concentration of the final solution.

How many grams of \(\mathrm{NaCl}\) are required to precipitate most of the \(\mathrm{Ag}^{+}\) ions from \(2.50 \times 10^{2} \mathrm{~mL}\) of \(0.0113 M\) AgNO \(_{3}\) solution? Write the net ionic equation for the reaction.
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