Question

A 12.9 g sample of an aqueous solution of perchloric acid contains an unknown amount of the acid. If 10.7 mL of 0.338 M barium hydroxide are required to neutralize the perchloric acid, what is the percent by mass of perchloric acid in the mixture

Answer 1

Answer:

$\% HClO_4=5.65\%$

Explanation:

Hello,

In this case, since the reaction between barium hydroxide and perchloric acid is:

$2HClO_4+Ba(OH)_2\rightarrow Ba(ClO_4)_2+2H_2O$

It means there is a 2:1 mole ratio between the acid and the base; thus we compute the moles of barium hydroxide that are reacting:

$n_{Ba(OH)_2}=0.0107L*0.338 mol/L=0.00363molBa(OH)_2$

Now, we compute the mass of perchloric acid (molar mass = 100.46 g/mol) by considering that 2:1 mole ratio:

$m_{HClO_4}=0.00363molBa(OH)_2*\frac{2molHClO_4}{1molBa(OH)_2} *\frac{100.45gHClO_4}{1molHClO_4} =0.729gHClO_4$

Finally, the percent of perchloric acid in such sample is:

$\% HClO_4=\frac{0.729g}{12.9g} *100\%\\\\\% HClO_4=5.65\%$

Best regards!