Answer:
30.12 mL.
Step-by-step explanation:
We'll begin by calculating the molarity of the phosphoric acid. This can be obtained as follow:
Phosphoric acid H3PO4 will dissociate in water as follow:
H3PO4(aq) <==> 3H^2+(aq) + PO4^3-(aq)
From the balanced equation above,
1 mole of H3PO4 produces 3 moles of H+.
Therefore, XM H3PO4 will produce 8.70 M H+ i.e
XM H3PO4 = 8.70/3
XM H3PO4 = 2.9 M.
Therefore, the molarity of the acid solution, H3PO4 is 2.9 M.
Next, we shall write the balanced equation for the reaction. This is illustrated below:
2H3PO4 + 3Ba(OH)2 —> Ba3(PO4)2 + 6H2O
From the balanced equation above, we obtained the following:
Mole ratio of the acid, H3PO4 (nA) = 2
Mole ratio of the base, Ba(OH)2 (nB) = 3
Data obtained from the question include the following:
Molarity of base, Ba(OH)2 (Mb) = 6.50 M
Volume of base, Ba(OH)2 (Vb) =.?
Molarity of acid, H3PO4 (Ma) = 2.9 M
Volume of acid, H3PO4 (Va) = 45 mL
The volume of the base, Ba(OH)2 Needed for the reaction can be obtained as follow:
MaVa /MbVb = nA/nB
2.9 x 45 / 6.5 x Vb = 2/3
Cross multiply
2 x 6.5 x Vb = 2.9 x 45 x 3
Divide both side by 2 x 6.5
Vb = (2.9 x 45 x 3) /(2 x 6.5)
Vb = 30.12 mL
Therefore, the volume of the base, Ba(OH)2 needed for the reaction is 30.12 mL