Question

1. Describe how you would prepare 1.50 L of a 0.500 M phosphate buffer with a pH of 7.5. (show work)

2. If 5.00 mL of 0.150 M HCl is added to 0.750 L of your buffer from (1), what will the new ph be? (show work)

3. If 10.00 mL of 1.50 M NaOH is added to 0.750 L of your buffer from (1), what will the new ph be? (show work)

Answer 1

Q1.

for a buffer with pH = 7.5 ,the best solution is pKa2 since pH = 7.21 anwe must be near 7.5

so

NaH2PO4 --> Na+ + H2PO4-

Na2HPO4 --> 2Na+ + HPO4-

there is present of H2PO4- (acid) and HPO4- (conjugate base)

therefore:

pH = pKa2 + log(HPO4-/H2PO4-)

7.5= 7.21 + log(ratio)

ratio = 10^(7.5-7.21) = 1.949

HPO4-/H2PO4- = 1.949

[Na2HPO4] = 1.949*[NaH2PO4 ]

so

[Na2HPO4] + [NaH2PO4 ] = 0.5

then

1.949*[NaH2PO4 ]+ [NaH2PO4 ] = 0.5

2.949[NaH2PO4 ] = 0.5

[NaH2PO4 ] = 0.5/2.949 = 0.1695 M

so..

[Na2HPO4] = 1.949*[NaH2PO4 ] = 1.949*0.1695 = 0.330355 M

so prepare the solution and mix both

in order to form V = 1.5 L buffer

Q2.

mmol o fHCL = MV = 5*10.15 = 50.75 mmol of HCl = 50.75*10^-3 = 0.05075mol of HCl

added to 0.75 L of buffer...

mmol of acid = MV = 0.75*0.1695 = 0.127125 mol of acid

mmol of base = MV = 0.75*0.330355 = 0.24776 mmol of base

after reaction of HCL.

mmol of acid = 0.127125 - 0.05075 = 0.177875

mmol of base = 0.24776 + 0.05075 = 0.29851

new pH

pH = pKa2 + log(HPO4-/H2PO4-)

pH = 7.21+ log(0.29851/0.177875) = 7.434

Q3

for NAOH

mmol = MV = 10*1.5 = 15 mmol = 15*10^-3 = 0.015 mol

mmol of acid = 0.127125 - 0.015   = 0.112125

mmol of base = 0.24776 +0.015 = 0.26276

pH = 7.21+ log(0.26276/0.112125 ) = 7.57985