HCOOH/NaHCOO Read Answer Items for Question 1 NaH 2 PO 3 /Na 2 HPO 3 Read Answer
ID: 797806 • Letter: H
Question
HCOOH/NaHCOO Read Answer Items for Question 1
NaH2PO3/Na2HPO3 Read Answer Items for Question 1
HOBr/NaOBr Read Answer Items for Question 1 Answer A. [H3O+] = 10-4 B. [H3O+] = 10-7 C. [H3O+] = 10-9
- 2-Which indicator would be the best to use for a titration between 0.30 M C6H5COOH with 0.30 M NaOH? You will probably need to consult the appropriate table in the book.
1. methyl red, color change at pH 3-5
2. bromothymol blue, color change at pH 6-8
3. crystal violet, color change at pH 0-2
4. phenolphthalein, color change at pH 8-10
5. alizarin yellow R, color change at pH 10-12
Explanation / Answer
(A) pH = -log[H3O+] = -log(10^(-4)) = 4.00
Ka of HCOOH = 1.8 x 10^(-4)
pKa of HCOOH = -log Ka = -log(1.8 x 10^(-4)) = 3.74
Thus HCOOH/NaHCOO is the best acid base pair for [H3O+] = 10^(-4) M as pKa of 3.74 is closest to desired pH of 4.00
(B) (A) pH = -log[H3O+] = -log(10^(-7)) = 7.00
Ka of H2PO3- = 2.0 x 10^(-7)
pKa of H2PO3- = -log Ka = -log(2.0 x 10^(-7)) = 6.70
Thus NaH2PO3/Na2HPO3 is the best acid base pair as pKa for [H3O+] = 10^(-7) M of 6.70 is closest to desired pH of 7.00
(C) (A) pH = -log[H3O+] = -log(10^(-9)) = 9.00
Ka of HOBr = 2.0 x 10^(-9)
pKa of HOBr = -log Ka = -log(2.0 x 10^(-9)) = 8.70
Thus HOBr/NaOBr is the best acid base pair for [H3O+] = 10^(-9) M as pKa of 8.70 is closest to desired pH of 9.00
C6H5COOH + NaOH => C6H5COONa + H2O
At the equivalence point:
[C6H5COO-] = 0.30/2 = 0.15 M (since volume is doubled)
C6H5COO- + H2O <=> C6H5COOH + OH-
Kb = [C6H5COOH][OH-]/[C6H5COO-] = [OH-]^2/[C6H5COO-]
= Kw/Ka = 10^(-14)/6.5 x 10^(-5) = 1.54 x 10^(-10)
[OH-] = (0.15 x 1.54 x 10^(-10)^(1/2) = 4.804 x 10^(-6) M
pOH = -log[OH-] = -log(4.804 x 10^(-6)) = 5.32
pH = 14 - pOH = 14 - 5.32 = 8.68
Thus the best indicator would be: (4) phenolphthalein, color change at pH 8-10
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