Question

What is the molarity of H₃PO₄ solution if 358 mL is completely titrated by 876 mL of 0.0102 M Ba(OH)₂ solution?

Answer 1

$H_ 3 P O_ 4 ( a q ) + B a ( O H )_ 2 ( a q ) → B a H P O_ 4 ( a q ) + 2 H_ 2 O ( l )$

Now moles of $Ba(OH)_2=0.0102 \times876 \times10^{-3} = 8.94 \times 10^{-3} moles$

As the ratio of $Ba(OH)_2$ and $H_3PO_4$ is 1:1 so the moles of $H_3PO_4=8.94 \times 10^{-3}$

Now Molarity of $H_3PO_4 =$ $\dfrac{moles}{volume(ml )}\times1000$

$=$ $\dfrac{8.94\times10^{-3}}{358} \times 1000=0.025M$