contestada

What is the molarity of an H3PO4 solution if 362 mL of the solution are completely titrated by 849 mL of 0.128 M Ba(OH)2 solution? (You will need to write a balanced chemical equation because this titration is NOT 1:1.)

0.450 M

0.200 M

0.300 M

0.0364 M

Respuesta :

Chlidonias

The balanced chemical equation between [tex] H_{3}PO_{4} and Ba(OH)_{2} [/tex] :

[tex] 2H_{3}PO_{4}(aq) + 3Ba(OH)_{2}(aq) -->Ba_{3}(PO_{4})_{2}(aq)+ 6H_{2}O(l) [/tex]

Moles of [tex] Ba(OH)_{2} =849 mL * \frac{1L}{1000mL} * \frac{0.128 mol}{L} = 0.108672 mol [/tex]

Moles of [tex] H_{3}PO_{4} [/tex] = [tex] 0.108672 mol Ba(OH)_{2} * \frac{ 2 molH_{3}PO_{4}}{3 mol Ba(OH)_{2}} = 0.072448 mol H_{3}PO_{4} [/tex]

Molarity of [tex] H_{3}PO_{4} = \frac{0.072448 mol}{(362 mL * \frac{1 L}{1000mL})}
= 0.200 M [/tex]

Correct answer is 0.200 M

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