Phenolphthalein would not be an appropriate indicator to use to determine K, for
ID: 578435 • Letter: P
Question
Phenolphthalein would not be an appropriate indicator to use to determine K, for phosphoric acid. Why not? Choose a suitable indicator from the following color chart. pH Indicator Phenolphthalein Methyl Red Orange 5. What would be the pH of a solution prepared by combining equal quantities of NaH,PO, and Na,HPO 1 23 4 5 67 8 910 11 Colorless Pink Red Red Orange Yellow Orange Peach Yellow Explain with an equation. 6. Sufficient strong acid is added to a solution containing Na,HPO, to neutralize one-half of t. What will be the pH of this solution? Explain.Explanation / Answer
4)
First equivalent point pH is calculated as follows
pH = 1/2(pKa1 + pKa2) = 1/2(2.12 + 7.21) = 4.67
so, the fisrst equivalence point pH is 4.67
phenolpthalein is not changing its colour arruond this pH, so it will not be a appropriate indicator
Methyl red change its color at this pH range
So, methyl red is suitable indicator
5)
Henderson-Hasselbalch equation is
pH = pKa + log([A-] /[HA])
Second dissociation of phosphoric acid is
H2PO4- <--------> HPO42- + H+
Ka2= [HPO42-] [H+] /[H2PO4-] = 6.2×10-8
pKa2 = 7.21
therefore,
when [H2PO4-] = [HPO42-]
log([H2PO4-]/[HPO42-] ) = 0
Then
pH = pKa
Therefore,
pH =7.21
6) pH = pKa + log([A-] /[ HA] )
= pKa2 + log(HPO42-]/[H2PO4-])
= 7.21 + 0
= 7.21
Picture 2
1) H3PO4(aq) + H2O(l) - - - - - - > H2PO4-(aq) + H3O+(aq)
2) Ka1 = [H2PO4-] [H3O+] /[H3PO4]
3) pKa1 = 2.12
Applying Henderson - Hasselbalch equation
pH = pKa + log([A-] /[HA])
= 2.12 + log([H2PO4-] /[H3PO4])
when [H2PO4-] = [H3PO4]
log([H2PO4-] /[H3PO4]) = 0
then
pH =pKa
Therefore,
pH = 2.12