15. The iodine liberated by the addition of an excess of KI and H2SO4 to 25.00 m

Question

15. The iodine liberated by the addition of an excess of KI and H2SO4 to 25.00 mL of 0.0236 M KIO required 23.55 mL of a solution for Na S2Os for titration using starch as the indicator What is the molarity of the Na2S2Os solution? 16. Balance the following equation then calculate the calculate the molarity of a hydrochloric acid solution requiring 0.1500 g of Na2COs for titration to the methyl orange end point: Na2C20s(aq) + HCI(aq) NaCl(aq)+ CO,(g) +H20(0) 17. A 0.1050 g sample of a divalent metal oxide, MO, was dissolved in 25.00 mL of a 0.3000 M hydrochloric acid (an excess), and the unreacted acid required 13.00 mL of a 0.1000 M NaOH for titration to the phenolphthalein end point. What is the molar mass and formula of the oxide? MO(s) +2 HCl(aq) MCl2(aq) +H20() 2

Explanation / Answer

15.

Moles of KIO3 = molarity * volume in L

n = 0.0236 * 25.00 / 1000

n = 0.000590 mol

From the first balanced equation,

2 mol of KIO3 forms 3 mol of I2

Then,

0.000590 mol of KIO3 forms 3 * 0.000590 / 2 = 0.000885 mol of KI

From the second balanced equation,

1 mol of I2 requires 2 mol of Na2S2O3

then, 0.000885 mol of I2 requires 2 * 0.000885 = 0.00177 mol of Na2S2O3

Therefore,

Molarity of Na2S2O3 solution = moles / Volume in L

M = 0.00177 / 0.02355

M = Molarity of Na2S2O3 = 0.0752 M

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