A 13.4 g sample of an aqueous solution of hydroiodic acid contains an unknown am

Question

A 13.4 g sample of an aqueous solution of hydroiodic acid contains an unknown amount of the acid. If 13.8 mL of 1.04 M potassium hydroxide are required to neutralize the hydroiodic acid, what is the percent by mass of hydroiodic acid in the mixture? % by mass A 8.38 g sample of an aqueous solution of nitric acid contains an unknown amount of the acid. If 29.4 mL of 0.188 M barium hydroxide are required to neutralize the nitric acid, what is the percent by mass of nitric acid in the mixture? % by mass

Explanation / Answer

a) 13.8 mL 1.04 M NaOH is used to neutralize HI in the sample.

Number of moles of NaOH used = (1.04/1000) *13.8 = 0.0143

(note: Molarity is the number of moles of solute in 1000 mL solution)

HI + NaOH gives NaI + H2O

from this equation, it is clear that one mole NaOH react with one mole HI

or 0.0143 moles HI is there in the unknown sample

molar mass of HI is 127.911 g/mole

mass of HI = number of moles x molar mass = 0.0143 x 127.911 = 1.8357 g

13.4 g sample contains 1.8357 g HI

% of HI = (1.8357/13.4)*100 =13.7 %

b) the reaction between barium hydroxide and nitric acid is

Ba(OH)2 + 2 HNO3 gives H2O + Ba(NO3)2

29.4 mL 0.188 M barium hydroxide is used for neutralization

Number of moles of Ba(OH)2 used = (0.188/1000) *29.4 = 0.0055 moles

From the balanced chemical equation, it is clear that 1 Ba(OH)2 reacts with 2 moles HNO3

thus the unknown sample contains = 2* 0.0055 = 0.011 moles of HNO3

molar mass of HNO3 is 63.01 g/mol

mass of HNO3 = 0.011 *63.01 =0.696 g

8.38 g contains 0.696 g HNO3

mass % of HNO3 = 0.696/8.38)*100 =8.3 %

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