1)Determine the molarity (in mol/L) of NH 4 + ions in solution when 3.60×10 -2 g

Question

1)Determine the molarity (in mol/L) of NH4+ ions in solution when 3.60×10-2 g of ammonium oxalate is added to enough water to prepare 5.23×102 mL of the solution

2)Calculate the molarity (in mol/L) of the IO3 ion in a solution resulting from the dilution of 73.9 mL of 3.88×10-1 M Ca(IO3)2 to a final volume of 687 mL. Report your answer to three significant figures.

3)Calculate the pH (to two decimal places) for the following:  5.67×10-1 M NaH

4)Determine the pOH (to two decimal places) of the solution that is produced by mixing 4.4 mL of 8.02×10-1 M K2O with 7.88 mL of 4.40×10-3 M Mg(OH)2.

5)A sample of tris(hyroxymethyl)aminomethane (H2NC(CH2OH)3), which is a base, with a mass of 0.5175 g is dissolved in water and used to standardize a solution of nitric acid. An endpoint is reached when 40.74 mL of nitric acid has been added. Determine the molarity (in mol/L) of the nitric acid solution. Report your answer to four significant figures.

6)The standardized nitric acid solution is then used to titrate 48.25 mL of a solution of sodium lactate (NaC2H5OCOO). An endpoint is reached when 15.22 mL of nitric acid has been added. Determine the molarity (in mol/L) of the sodium lactate solution. Report your answer to four significant figures.

Explanation / Answer

(1)

Moles of Ammonium oxalate = mass / molar mass = 3.60 * 10-2 / 124.1 = 2.90 * 10-4 mol

Volume of solution = 5.23 * 102 mL = 5.23 * 102 / 1000 L = 5.23 * 10-1 L

Molarity of ammonium oxalte = number of moles / volume of solution in L

= 2.90*10-4/(5.23*10-1) = 5.54 * 10-4 M

Dissociation of ammonium oxalate can be written as,

(NH4)2C2O4 (aq.) --------------> 2 NH4+ (aq.) + C2O42- (aq.)

5.54*10-4                                      2*5.54*10-4

Therefore, Molarity of ammonium ions = 1.108 * 10-3 M

(2)

Dilution formula,

M1V1 = M2V2

3.88 * 10-1 * 73.9 = M2 * 687

M2 = Molarity of Ca(IO3)2 = 4.17 * 10-2 M

Dissociation of Ca(IO3)2 can be written as,

Ca(IO3)2 (aq.) --------------> Ca2+ (aq.) + 2 IO3- (aq.)

4.17*10-2                                                    2*4.17*10-2

Therefore, molarity of IO3- ions = 8.34 * 10-2 M

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