Question

A formic acid buffer contains 0.45 M HCOOH and 0.40 M HCOONa. Strong base (10.00 mL of 1.00 M NaOH) is added to 100.0 mL of the buffer. After the addition, what is the [HCOO^—] in the buffer?

0.35 M
0.50 M
0.050 M
0.45 M
0.25 M

The answer is .45 but can you show all the work to get this answer

Answer 1

Molarity = Number of moles of solute per liter of solution

Nummber of moles of HCOO^- = (0.40mol/1000ml) × 100ml = 0.040mol

added base reacts with weak acid HCOOH

OH^- + HCOOH -------> HCOO^- + H2O

number of moles of NaOH added = ( 1mol/1000ml)×10ml = 0.01mol

0.01mol of NaOH react with 0.01moles of HCOOH to give 0.01moles of HCOO^-

After addition of NaOH

Total volume = 100ml + 10ml = 110ml

moles of HCOO^- = 0.040mol + 0.010mol = 0.050ml

[HCOO^-] = (0.050mol/110ml )×1000ml = 0.45M