A 500-mg tablet containing antacids plus inert material was dissolved in 50.0 mL

Question

A 500-mg tablet containing antacids plus inert material was dissolved in 50.0 mL of 0. 500 M HC1. The resulting solution required 26.5 mL of 0.377 M NaOH for neutralization. How many moles of OH^- were in the tablet? If the tablet contained equal masses of Al(OH)_3 and Mg(OH)_2, how many moles of each hydroxide were present? The volume of the oceans is estimated to be 1.37 times 10^9 km^3. The concentration of gold in sea-water is about 4 times 10^-6 mg per liter of sea-water. Estimate the total mass of gold dissolved in the oceans. A 0.7755-g portion of a solid mixture containing sodium hydroxide and unreactive impurities is dissolved in water and titrated with standard 0.1000 M H_2SO_4. If 34.44 mL of the acid is required to neutralize the sample, calculate the percent by weight of NaOH in the mixture.

Explanation / Answer

a)moles of HCl in 50ml of 0.5M= 0.5*50/1000=0.025 moles

moles of NaOH in 26.5 ml of 0.377M= 0.377*26.5/1000 =0.0099

millimoles of OH in the tablet are neutralized by the acid accordingly

HCL+ NaOH----> NaCl + H2O

So 1 mole of HCl requires 1 mole of NaOH.

So moles of OH present in the tablet = 0.025-0.0099 = 0.0151 moles

b) the reactino with HCl are Al(OH)3 + 3HCl -----à AlCl3 + 3H2O and Mg(OH)2+ 2HCl-àMgCl2+ 2H2O

let x= mass of Al(OH)3, molar mass of   Al(OH)3= 78, moles of Al(OH)3= x/78

Al(OH)3------> Al+ + 3OH- moles of OH- = 3x/78

  Since x= mass of A Mg(OH)2, molar mass of Mg(OH)2= 58, moles of Mg(OH)2= x/58

Mg(OH)2-> Mg+2 +2OH-, moles of OH- = 2x/58

So total molex x*(3/78+2/58)= 0.0151,x =0.21

Moles of OH- in Al(OH)3= 3*0.21/78 =0.0080 and moles of OH- in Mg(OH)2= 2*0.21/58= 0.0072

c)

1Km= 1000m =1000*100cm= 105cm , 1KM3= 1015cm3 =1015/1000 L ( 1000cc =1L) =1012 L

sea water contains 4*10-6 mg/L, total gold = 4*10-6*1012 mg =4*106 mg = 4*1000gm

d)

The reaction between NaOH and H2SO4 is

2NaOH+ H2SO4-->Na2SO4 + 2H2O, 1 mole of H2SO4 requires 2 moles of H2SO4.

Moles of H2SO4 in 34.44ml of 0.1M= 0.1*34.4/1000 =0.00344 moles

Hence moles of NaOH required =0.00344*2= 0.00688, Molar mass of NaOH= 40

Mass of NaOH in the sample = 0.00688*40= 0.2752 gm

Mass % of NaOH =100*0.2752/0.7755=35.5%

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