A student add a total of 15.00 m L of .5432 HCl to their unknownalkaline earth c
ID: 75851 • Letter: A
Question
A student add a total of 15.00 m L of .5432 HCl to their unknownalkaline earth carbonate .it completely dissolved and stoppedfizzing . Addition of 3 drops of bromophenol blue confirmed the solution was acidic . the solutionrequired 12.3 m L of 12.34 m L of 0.1255 M NaOH to back titrate theexcess HCl.
1) What was the total number ofmoles of HCl add to the unknown ?
2) How many moles of excess HClwere added?
3) How many moles of MCO3 werepresent in the original unknown ?
4) If the mass of the originalunknown was .04871 g , and the unknown was 100% MCO3 . what is theformula mass of the carbonate ?
5) Identify unknown ?
Explanation / Answer
1) Total moles of HCl added = molarity * volume = 0.5432 M * 0.015 L =0.008148 moles . 2) The excess moles of HCl were back titrated with NaOH . Moles of NaOH = 0.1255 M * 0.01234 L =0.001548 moles . So excess moles of HCl = 0.001548 moles. . 3) Since it is an alkaline earth carbonate, general formula isMCO3. i.e. one mole of carbonate ion per mole of carbonatesalt. Hence moles of carbonate = moles of HCl consumed. . Moles of HCl consumed = Total moles - excess moles =0.008148 -0.001548 =0.0066 moles = moles of MCO3in the unknown . 4) Formula mass of unknown = mass / moles = 0.04871 g / 0.0066 moles =7.38 g/mol . 5) The answer is part 4 is highly unlikely. So please check yourvalues if there is any mistake in decimal points or values. Theunknown can be identified only from correct formula mass. . 4) Formula mass of unknown = mass / moles = 0.04871 g / 0.0066 moles =7.38 g/mol . 5) The answer is part 4 is highly unlikely. So please check yourvalues if there is any mistake in decimal points or values. Theunknown can be identified only from correct formula mass.Related Questions
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