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

`\% HClO_4=5.65\%`

Step-by-step 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*(2molHClO_4)/(1molBa(OH)_2) *(100.45gHClO_4)/(1molHClO_4) =0.729gHClO_4`

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

`\% HClO_4=(0.729g)/(12.9g) *100\%\\\\\% HClO_4=5.65\%`

Best regards!