CHEMISTRY please help: buffer- titration- acid/base questions

Question

okay so i know these maybe a little easier than i think but i keep coming out with numbers that are not even possible answers i was just wondering if someone could please show me/ talk me through the steps to do these. My exam is in 4 days Its just a few review:

1) An acetic acid buffer containing 0.50 M CH3COOH and 0.50 M CH3COONa has a pH of 4.74. What will the pH be after 0.0020 mol of HCl has been added to 100.0 mL of the buffer?

a. 4.77

b. 4.71

c. 4.68

d. 4.62

e. none of the above

--> im getting an answer thats not on here but the closest i get is 4.65 which my guess will be eiether c or d

2)

A 20.0-mL sample of 0.25 M HNO3 is titrated with 0.15 M Na OH. What is the pH of the solution after 30.0 mL of NaOH have been added to the acid?

a. 2.00

b. 1.60

c. 1.05

d. 1.00

e. none of the above

--->1.6??????? please double check this for me

3)

A buffer is prepared by adding 150 mL of 1.0 M NaOH to 250 mL of 1.0 M NaH2PO4. How many moles of HCl must be added to this buffer solution to change the pH by 0.18 units?

a. 0.025 mol HCl

b. 0.063 mol HCl

c. 0.082 mol HCl

d. 0.50 mol HCl

e. 1.0 mol HCl

i actually have no idea how to do this one.

id really appreciate your help

Answer 1

since you have equal amounts of the acid and base forms to start with and the pH is 4.74  then 4.74 is the pKa  because

pH = pKa + log base/acid  log 1 = 0,  so pH = pKa

100 mL of the buffer contains 0.5 X 100 = 50 mmoles of both the acid and the base  so when you add 0.002 moles of acid  that is 2 mmoles..you will react that 2 mmoles of acid with 2 mmoles of the base form and convert it to 2 mmoles of the acid   soo the pH will now be  pH = 4.74 + log 48mmol/52mmol = 4.71 = b

20 mL of 0.25 M HNO3 = 20 x 0.25 = 5 mmol of H+

30 mL of 0.15 M NaOH = 30 X 0.15 = 4.5 mmole OH-

these two react leaving 0.5 mmoles of H+ in the total of 50 mL this = 0.5 /50 mL = 0.01M  = 10^-2 M in H+   pH = 2 = a

to change the pH by 0.18 units when adding an acid..since the pKa is positive  the only way to get that number to get smaller ( more aidic ) is to have the units -.18  ..so the log of base /acid = - 0.18 so the log inverse of that is 0.667 or 2/3    so the ratio of the base to the acid must be 2/3    orr 100/150 ( since it must total to 250 mmoles total )

soo if you start with  150 mmoles of base and 100 mmoles of acid then adding 50 mmoles of acid will reduce the base by 50 mmoles and increase the acid by 50 mmoles ...to add 50 mmoles = 0.05moles H+

Th estart values are obtained by 250 mL x 1M = 250 mmoles reacted with 150 x 1 M NaOH yields 150 mmoles of the base form ( HPO4^-2 and 100 mmoles of the acid form remaining

Answer 2

wait so the answer to the third question would be 1.0 HCL then?