A buffer is prepared by adding 150 mL of 1.0 M NaOH to 250 mL of...
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?
The answer is .025mol
Please use chart if possible. Show all steps and calulations please
Homework Answers
The pH of a buffer can be calculated as
pH = pKa + log([A-]/[HA])
Your buffer is made up of H2PO4-
and HPO42- , because you partially
neutralized your acid (H2PO4-)
with NaOH.
So,
pKa = 7.20 (second ionization of
H3PO4)
HA = H2PO4-
A- = HPO42-
You had 0.25 moles of your HA and added 0.15 moles of NaOH.
Now you have 0.10 moles HA and 0.15 moles A-.
The ratio of concentrations ([A-]/[HA]) is the ratio of
moles 0.15/0.10 = 1.5
pH = 7.20 + log(1.5) = 7.20 + 0.18 = 7.38
You want
pH = 7.20 + log(1.0) = 7.20
That means you want half of your phosphate in each form:
0.125 moles of HA and 0.125 moles of A-.You need to
convert 0.025 moles of A- to HA by adding 0.025
moles HCl.
**************
reacting 150mL of 1M NaOH ( 150 mmoles OH-1 ) with 250 mLof
1M H2PO4^-1 ( 250 mmoles ) results in production of 150 mmoles of the base form ( HPO4^-2 ) and leaves 100 mmoles of the acid form.
So the initial ratio of th 2 forms is
= 150base/100 acid.
To change the pH 0.18 units by adding acid this means that the pH must go down by 0.18 units so in the H/H relationship
where pH = pKa + logbase/acid
the log term must be - 0.18 the inverse log for -0.18 = 0.6667 = a ratio of base to acid of 2/3....so reacting the base form with 50 mmoles of HCl will reduce the base form by 50 mmoles and add 50 mmoles to the acid form making a ration of 100 base /150 acid = 2/3
so the answer should be 0.025 moles of HCL
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The pH of a buffer can be calculated as
pH = pKa + log([A-]/[HA])
Your buffer is made up of H2PO4- and HPO42- , because you partially neutralized your acid (H2PO4-) with NaOH.
So,
pKa = 7.20 (second ionization of H3PO4)
HA = H2PO4-
A- = HPO42-
You had 0.25 moles of your HA and added 0.15 moles of NaOH.
Now you have 0.10 moles HA and 0.15 moles A-.
The ratio of concentrations ([A-]/[HA]) is the ratio of moles 0.15/0.10 = 1.5
pH = 7.20 + log(1.5) = 7.20 + 0.18 = 7.38
You want
pH = 7.20 + log(1.0) = 7.20
That means you want half of your phosphate in each form:
0.125 moles of HA and 0.125 moles of A-.You need to convert 0.025 moles of A- to HA by adding 0.025 moles HCl.