Question

how do I solve this

1. (10 pts) 9.00ml of a 0.300M solution of lead (I) nitrate, Pb(NO3), is added to 9.50ml of a 0.200M solution of aluminum chloride, AICI. The unbalanced equation is shown below. What is the mass of lead (1) chloride, PbCl2, is it possible to produce in this reaction? (first determine which reactont is limiting). Pb(NO3)2(aq) + AICI: (aq) + PbCl2(S) + AI(NO3)3(aq) 2. (5 pts) A 1.00M solution of Ba(OH) is used to titrate a solution of phosphoric acid, H,PO, of unknown concentration. At the end of the titration it's found that 42.00 ml of the base, Ba(OH)2, and 40.00 ml of phosphoric acid, H.PO, were used to reach the endpoint. What is the concentration of the phosphoric acid solution? The balanced equation for the reaction is: 3Ba(OH)2 + 2H, PO, 6H2O + Ba,(PO).

solve question 2 only, thank you

Answer 1

To find the concentration of Phosphoric acid titrated with Ba(OH)2

we just use simple expression:-. M​​​​​​1V1=M2V2 but, here we have given the balanced equation of the reaction from which we can get information that for 1 mole of H3PO4 we need 3/2 moles of Ba(OH)2 were used.

So, now the equation becomes 2CaVa=3CbVb

Where Ca is conc.of phosphoric acid, Va is volume of phosphoric acid, Cb is conc. Of Ba(OH)2 and Vb is volume of Ba(OH)2 used.

From the above given information we just put values into the above equation

Ca = (3/2) * {1.00* 42* 10^-3} /40*10^-3

Ca= 1.574 M

So, the concentration of Phosphoric acid used is 1.574 M