posted by Anonymous .
A titration involves titrating 0.4 M NaOH into a solution of 0.2 M H3PO4.
a.) Calculate the volume of NaOH that will be required to reach the first equivalence point.
b.) Calculate the volume of NaOH required to reach the second equivalence point.
Other information: 10 ml H3PO4 was added to 250 ml of deionized water, and solution was then titrated with NaOH.
Is that 10 mL of the 0.2M H3PO4 added to 250 mL H2O to make 260 mL solution? If you titrate the entire solution, which seems unlikely, then you will have
H3PO4 ==> NaH2PO4 + H2O
mols H3PO4 = 0.010L x 0.2M = 0.002
mols NaOH required = 0.002 since 1 mol H3PO4 = 1 mol NaOH.
Then M NaOH = mols NaOH/L NaOH or
L NaOH = mols NaOH/M NaOH = 0.002/0.4 = approx 0.005L or 5 mL
For part B you are titrating the second H do it will take twice that or abouat 10 mL.
If I've interpreter the problem incorrectly please rephrase the question so it is clear what is being titrated.
Oh, sorry. It was supposed to be 50 ml of water instead of 250 ml.