A 0.5 g sample of dry soda ash is dissolved in 50 mL of distilled water. Phenolp

Question

A 0.5 g sample of dry soda ash is dissolved in 50 mL of distilled water. Phenolphthalein is added and 15.00 mL of 0.1000 M HCI is required to reach the FIRST end point. Excess volume of HCl is added according to our manual. After boiling, the excess HCl is titrate with 0.1000 M NaOH Finally, 13.50 mL of the NaOH is required to reach the end point of titration a mass percentage of carbonate in the soda sample? (don't forget to calculate how much acid on earth you used. Do you know the mass of sodium carbonate? Also remember the molar ratio of Na_2CO_3 and HCl)

Explanation / Answer

The reaction we have is

Na2CO3 + 2HCl --> 2NaCl + H2O + CO2

Initial HCl moles used = M x V ( inL) = 0.1 x ( 0.015) = 0.0015              ( where V = 15 ml = 0.015L)

NaOH moles titrated = M x V ( of NaOH) = 0.1 x ( 0.0135) = 0.00135

since HCl and NaoH react in 1:1 ratio the excess HCl moles which is titrated with NaoH = NaOH moles

                    = 0.00135

HCl moles initial = HCl moles titrated with Na2CO3 + HCl moles titrated with NaOH

0.0015 = HCl moles titrated with Na2CO3 + 0.00135

HCl moles used to titrate Na2CO3 = 0.00015

by balanced equation at starting the Na2CO3 moles = ( 1/20 HCl moles = ( 1/2) 0.00015 = 7.5 x 10^ -5

Na2CO3 mass = moles x molar mass of Na2CO3

                   = 7.5 x 10^ -5 mol x 106 g/mol

              = 0.00795 g

mass % of carbonate in soda sample = ( 100 x Na2CO3 mass / sample mass)

               = ( 100 x 0.00795 /0.5) = 1.59 %

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