Need Help! Write the balanced chemical equation for the reaction of KHP with NaO

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Need Help!

Write the balanced chemical equation for the reaction of KHP with NaOH. Suppose your laboratory instructor inadvertently gave you a sample of KHP contaminated with NaCl to standardize your NaOH. How would this affect the molarity you calculated for your NaOH solution? Justify your answer. A solution of malonic acid, H2C3H2O4, was standardized by titration with 0.100 M NaOH solution. If 21.82 mL of the NaOH solution were required to neutralize completely 12.12 mL of the malonic acid solution, what is the molarity of the malonic acid solution? H2C3H2O4 + 2NaOH rightarrow Na2C3H2O4 + 2H2O Sodium carbonate is a reagent that may be used to standardize acids in the same way that you have used KHP in this experiment. In such a standardization it was found that a 0.432-g sample of sodium carbonate required 22.3 mL of a sulfuric acid solution to reach the end point for the reaction. Na2CO3(aq) + H2SO4(aq) rightarrow H2O(l) + CO2(g) + Na2SO4(aq) What is the molarity of the H2SO4? A solution contains 0.252 g of oxalic acid, H2C2O4. 2H2O, in 500 mL. What is the molarity of this solution?

Explanation / Answer

1.Potassium acid phthalate is a large molecule (KHC8H4O4) with a molar mass of 204.2 g/mol. Instead of writing the whole formula, we abbreviate it as KHP, where "P" stands for the phthalate ion, (C8H4O4)- - and not for phosphorus. KHP is an acidic substance, with the ionizing hydrogen being set forward in the formula for emphasis. Therefore, KHP is monoprotic and will react with NaOH in a simple 1 to 1 relationship according to the following equation:

NaOH(aq) + KHC8H4O4 (aq) ---> KNaC8H4O4 (aq) + H2O(l)

2.Assuming you mean potassium hydrogen phosphate, K2HPO4, the net ionic equation would be:

(HPO4)- + (OH)- --> H2O + (PO4)3-
with (HPO4)- and (OH)- reacting in a 1:1 ratio.
Contamination with NaCl would mean that you actually weighed out less K2HPO4 than you thought. This would mean that the K2HPO4 solution was less concentrated than you thought. This would mean that a given volume of this solution (say 25cm3) would neutralise less NaOH than it should. Given that the concentration of NaOH would be calculated as:
[NaOH] = (assumed moles of K2HPO4)/(volume of NaOH(aq))
and 'volume of NaOH' is too small, the value you would calculate would be too large

3.Malonic acid (HOOCCH2COOH) has 2 replaceable protons

H2A + 2NaOH --> Na2A + 2H2O
You require 2 moles of NaOH for every 1 mole of malonic acid
21.82 mL NaOH x 0.1000 mole/L = 2.182 millimoles of NaOH (0.002076 mole)
Millimoles of malonic acid neutralized by 2.182 millimolesd of NaOH = 1.091 millimoles (2.182 / 2)
13.15 mL malonic acid x molarity = 1.091 millimoles
molarity = 1.091 millimoles / 12.12 mL = 0.0900 M (moles/L or millimoles/mL)
ANSWER: 0.0900 M

4.First find the number of moles of sodium carbonate used.

Moles = grams sodium carbonate over grams sodium carbonate per mole

Moles = .432 g over l06 g/mole sodium carbonate = 4.08 x l0^-3

Look at the formula for sodium carbonate and see that one mole of sodium carbonate will contain one mole of CO3 = ion, and since this ion has a charge of -2, it can neutralize two H+ ions.

Therefore 4.08 x l0^-3 moles of CO3= will neutralize twice as many moles of H+ or

8.15 x l0^-3 moles of H+

And in one mole of H2SO4 there are two moles of H+

So to get moles of H2SO4 neutralized, we cut moles H+ in half to get 4.08 x l0^-3 moles sulfuric acid neutralized.

Now Molarity sulfuric acid = moles acid over LITERS solution

so Molarity = 4.08 x l0^-3 moles over .0223 LITERS (same as 22.3 ml)

and Molarity of acid = l.83 Molar H2SO4

5.oxalic acid has molecular weight = 90.03 g/mole

0.252 g x (1 mole / 90.03 g) x ( 1 / 0.500L) = 0.0056 moles /L = 0.0056 M

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