I am doing a post lab for a acid and base buffers lab. I was asked to find the t
ID: 552341 • Letter: I
Question
I am doing a post lab for a acid and base buffers lab. I was asked to find the theoretical pH (what the pH should actually be) for each of my pH's in my data please help it is due tonight and that is the only question I cannot figure out. I just need one of the theoretical pHsd' one and I can do the rest.
Background:
30 mL of sodium dihydrogen phosphate and 20 mL k2HPO4 was added to buffer one
20 mL of sodium dihydrogen phosphate and 30 mL of k2HPO4 was added to buffer two
HCl .15 M
NaOH .15 M
.200 M K2HPO4
.200 M K2HPO4
01 mBuctex Lab Pariner Course Jordyn sonson corsne and Baxt veadivi nital uneer oaa 8.31 40-Som of NaDt Part p:Torturing me cord Buffer with Acid anal Lose 240 mu, 130-05 Date Instructor/TA Signature Date Student's Signature uNTAINH EAD PRES Note: Insert Foldover Back flap under each copy set before writing S 0040Explanation / Answer
Parts 1 and 2
We will need the pKa value of sodium dihydrogen phosphate, NaH2PO4 which is 7.20.
Write down the ionization reaction.
H2PO4- (aq) <======> H+ (aq) + HPO4- (aq); pKa = 7.20
H2PO4- is supplied by NaH2PO4 while K2HPO4 is the source of HPO42-. You haven’t mentioned the concentration of H2PO4- and HPO42-, but I shall assume them to be 0.200 M each.
Prepare the following table.
Part 1
Millimoles of H2PO4- = (volume of H2PO4-)*(concentration of H2PO4-)
Millimoles of HPO42- = (volume of HPO42-)*(concentration of HPO42-)
Total volume of solution (mL)
Concentration of H2PO4- = (millimoles of H2PO4-)/(total volume of solution)
Concentration of HPO42- = (millimoles of HPO42-)/(total volume of solution)
(30 mL)*(0.200 M)
(20 mL)*(0.200 M)
50
(30 mL)*(0.200 M)/(50 mL) = 0.12 M
(20 mL)*(0.200 M)/(50 mL) = 0.08 M
Part 2
(20 mL)*(0.200 M)
(30 mL)*(0.200 M)
50
(20 mL)*(0.200 M)/(50 mL) = 0.08 M
(30 mL)*(0.200 M)/(50 mL) = 0.12 M
Use the Henderson-Hasslebach equation:
pH = pKa + log [HPO42-]/[H2PO4-]
Plug in values and get
Part 1: pH = 7.20 + log (0.08 M)/(0.12 M) = 7.20 + log (0.6667) = 7.20 + (-0.1761) = 7.0239 7.02 (ans).
Part 2: pH = 7.20 + log (0.12 M)/(0.08 M) = 7.20 + log (1.50) = 7.20 + (0.1761) = 7.3761 7.40 (ans).
Part 3:
HCl dissociates completely into protons; the dissociation equation is
HCl (aq) -------> H+ (aq) + Cl- (aq)
As per the dissociation equation, [HCl] = [H+] = 0.15 M.
pH = -log [H+] = -log (0.15) = 0.8239 0.82 (ans).
Part 4:
NaOH dissociates completely into OH-; the dissociation equation is
NaOH (aq) --------> Na+ (aq) + OH- (aq)
As per the dissociation equation, [NaOH] = [OH-] = 0.15 M.
pOH = -log [OH-] = -log (0.15) = 0.8239 and pH = 14 – pOH = 14 – 0.8239 = 13.1761 13.18 (ans).
Part 1
Millimoles of H2PO4- = (volume of H2PO4-)*(concentration of H2PO4-)
Millimoles of HPO42- = (volume of HPO42-)*(concentration of HPO42-)
Total volume of solution (mL)
Concentration of H2PO4- = (millimoles of H2PO4-)/(total volume of solution)
Concentration of HPO42- = (millimoles of HPO42-)/(total volume of solution)
(30 mL)*(0.200 M)
(20 mL)*(0.200 M)
50
(30 mL)*(0.200 M)/(50 mL) = 0.12 M
(20 mL)*(0.200 M)/(50 mL) = 0.08 M
Part 2
(20 mL)*(0.200 M)
(30 mL)*(0.200 M)
50
(20 mL)*(0.200 M)/(50 mL) = 0.08 M
(30 mL)*(0.200 M)/(50 mL) = 0.12 M
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