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Problem 94 Hard Difficulty

Potassium acid phthalate, $\mathrm{KNaC}_{8} \mathrm{H}_{4} \mathrm{O}_{4},$ or $\mathrm{KHP}$ , is used in many laboratories, including general chemistry laboratories, to standardize solutions of base. KHP is one of only a few stable solid acids that can be dried by warming and weighed. A $0.3420-\mathrm{g}$ sample of $\mathrm{KNaC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}$ reacts with 35.73 $\mathrm{mL}$ of a NaOH solution in a titration.
What is the molar concentration of the NaOH?
$\mathrm{KNaC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}(a q)+\mathrm{NaOH}(a q) \longrightarrow \mathrm{K} \mathrm{NaC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(a q)$

Answer

0.04687$M$

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00:43

Sisi G.

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Chapter 4

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Video Transcript

to calculate the molar concentration of a solution. We need to know the moles of the solute and we need to know the volume of the solution. The salute in this case will be sodium hydroxide. So how do we get the moles of sodium hydroxide? Well we get the moles of sodium hydroxide from information associated with K. Hp. And the stock geometry of the reaction they both undergo. Historic geometry is 1 to 1. So if we have 0.34 20 g of K. H. P. This being K. H. P. We can convert those grams K HP into molds K HP. By dividing by the molar mass of K HP, which is 204.22 g per mole. Then we have malls K. H. P, recognizing the stock geometry is 1 to 1. This is also the moles of sodium hydroxide that were present in the solution that neutralized cage P. We then divide that by the volume of the sodium hydroxide solution, which is 35.73 million liters. But to get a molar concentration, we need to convert that to leaders. And this gives us a sodium hydroxide concentration of 0.4687 sodium hydroxide.

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Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson

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