Phenolphthalein would not be an appropriate indicator to use to determine K, for

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

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