0.8742 g of a metal carbonate (MCO 3 ) is added to 100 mL of water in a 250 mL b

Question

0.8742 g of a metal carbonate (MCO3) is added to 100 mL of water in a 250 mL beaker. To this, 20.00 mL of 1.8442 M HCl is added, which is in excess. The reaction is:

MCO3 (s) + 2 HCl (aq) --> M2+ + H2O + 2 Cl- + CO2

When the reaction is finished, the solution is quantitatively transferred to a 500.0 mL volumetric flask and diluted to the mark with water. This is solution A.

25.00 mL of solution A is titrated with a 0.03253 mol/L NaOH solution. The end point is 29.84 mL.

Determine the identity of the metal carbonate.

Explanation / Answer

first we find molarity of HCl IN FINAL SOLUTION

HCl + NaOH ------> Na^+ + cl^- H2O

molatity =(no. of moles )/(volume in liters)

no. of moles of NaOH used = no of moles of HCl used (for nuetralisation of rxn)

molarity of NaOH* volume in liters = molarity of HCl* volume used

0.03253* 29.84/(1000) = molarity of HCl * 25/(1000)

so,

Molarity of HCl in solution A=0.0387 --------------(1)

now we will find no of mole of HCl was present in the initial reaction:-

no of moles of Hcl=Molarity of Hcl * volume of HCl

=0.0387 * 500/(1000)

= 0.01935

now,

MCO3+ 2* HCl ---------->M^+2 +H2O +2*Cl- +CO2

for nuetralisation rxn

we need 2* mole of Hcl to nuetralise 1*mole of MCO3

so,

no of moles of HCl used in the rxn

=initial moles of HCL - moles used in rxn A

=20*1.8442/(1000) - 0.01935

=0.03688 - 0.01935

=0.01753

molesof HCl /2 = mole of MCO3 /1

0.01753/2 = moles of Mco3

Moles of MCO3 = 8.767 * 10^(-3)

=mass / molecular wt

=0.874/ molecular wt

molecular wt=100

= MCO3

=( M+12+16*3)

molecular wt of M=40

metal is Calcium

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