Use the HH equation.
pH = 4.80
pKa = 4.74
Solve for the ratio of (base)/(acid). That is equation 1.
The second equation is (base) + (acid) = 0.15
Solve these two equations for (base) and (acid), then convert to mL by using the dilution formula of
mL1 x M1 = mL2 x M2
mL1 x 1M acid = 50 mL x Mfrom problem.
Calculate amount of 1 M acetic acid and 1M sodium acetate necessary to prepare 50 ml of a 0.15 M buffer at ph 4.80. (pka acetic acid =4.74)
2 answers
pH = pKA + log salt/acid
4.80 = 4.74 + log salt/acid
4.80-4.74= log salt/acid
0.06 = log salt/acid
antilog0.06 = salt/acid
1.148 = salt/acid
total=1.148+1=2.148
salt=1.148*0.15/2.148 =0.080moles
acid = 1*0.15/2.148 = 0.0698moles
salt=moles*mol.wt=0.080*82.03=
6.56g
acid = moles*mol.wt=0.0698*60.05=
4.19g
1000ml of acid contains=4.19g of acid
1ml contains=4.19/1ooo
50ml contains=4.19/1000*50= 0.2095g per 50ml
1000ml of salt contains = 6.56g of salt
1ml of salt contains=6.56/1000
50ml of salt contains=6.56/1000*50=0.32g per 50ml
4.80 = 4.74 + log salt/acid
4.80-4.74= log salt/acid
0.06 = log salt/acid
antilog0.06 = salt/acid
1.148 = salt/acid
total=1.148+1=2.148
salt=1.148*0.15/2.148 =0.080moles
acid = 1*0.15/2.148 = 0.0698moles
salt=moles*mol.wt=0.080*82.03=
6.56g
acid = moles*mol.wt=0.0698*60.05=
4.19g
1000ml of acid contains=4.19g of acid
1ml contains=4.19/1ooo
50ml contains=4.19/1000*50= 0.2095g per 50ml
1000ml of salt contains = 6.56g of salt
1ml of salt contains=6.56/1000
50ml of salt contains=6.56/1000*50=0.32g per 50ml