Question

A 0.2726 g sample of metal was dissolved in 50.00 mL of 0.500 M HCl. After all the metal had dissolved, the leftover acid was titrated with 0.1054 M NaOH. If 24.36 mL of 0.1054 M NaOH were required to neutralize the leftover acid, what was the atomic mass of the metal? The metal dissolved to form M+2 ions in solution.

The correct answer is 24.3g, please show all your work, thank you.

Answer 1

The reaction between HCl and NaOH is

HCl(aq) + NaOH(aq ) --------> H2O(l) + NaCl(aq)

This reaction is 1:1 molar reaction

Molarity = Number of moles per liter of solution

Number of moles of NaOH consumed =(0.1054mol/1000ml)×24.36ml = 0.002568mol

0.002568moles of NaOH should react with 0.002568 moles of HCl

So, number of moles of HCl leftover = 0.002568mol

total number of moles of HCl added = (0.500mol/1000ml)×50ml = 0.025mol

Number of moles of HCl reacted = 0.0250mol - 0.002568mol = 0.002243mol

The reaction between HCl and the metal is as follows

2HCl(aq) + M ----------> MCl2 + H2

So, 0.002243moles of HCl should react with 0.002243mol/2 = 0.01122moles of M

mass of metal dissolved = 0.2726g

molar mass = Mass/number of moles

molar mass = 0.2726g/0.01122mol = 24.3g/mol

Therefore,

Atomic mass of the metal = 24.3g/mol