Need Help ASAP exam in a few days?

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Edit.

O.K., I see where the problem is, so I'll walk you through these.

(1) how many mL of 0.40 M H2SO4 would require the same amount of base to reach full neutralization as would 25 mL 0.60 M H3PO4?

a. 25.00 mL .40 M H2SO4

b. 37.50 mL .40 M H2SO4

c. 50.0 mL .40 M H2SO4

d. 56.25 mL .40 M H2SO4

We have to take the numbers of H+ that can be donated by each acid into consideration in our calculations.

H2SO4 => 2H+ + SO4^2-

H3PO4 => 3H+ + PO4^3-

(25 mL)*(0.60 M H3PO4)*(3H+/H3PO3) = 45 mmol of H+

(x mL)*(0.40 M H2SO4)* (2H+/H2SO4) = 45 mmol of H+

x mL = (45 mmol of H+)/((0.40 M H2SO4)*(2H+/H2SO4))

x mL = 56.25 mL

which is answer (d).

(2) Note : in flask two 40.0 mL of 0.30 N H3PO4 Question : What is the phosphate concentration in Flask 2 after 40 mL of 0.30 N NaOH has been added?

This is in essence a dilution problem. It is somewhat complicated by the use of Normality instead of Molarity for the units of concentration. I've included a reference on Normality for you to read to help you in understanding this concept.

We first convert Normality to Molarity.

By definition,

1M H3PO4 = 3N H3PO4

(molarity * number of H+ donated = normality)

So

0.30 N H3PO4 = 0.1 M H3PO4

Therefore,

Initial phosphate concentration = 0.1 M phosphate

Initial mmol of phosphate:

(40.0 mL)*(0.1 M phosphate) = 4.00 mmol of phosphate

This is diluted by adding another 40 mL of solution, so

Final volume = 40.0 mL + 40 mL = 80 mL

Final phosphate concentration = (4.00 mmol of phosphate)/(80 mL) = 0.050 M

which is answer (a).