Respuesta :
Answer:
(a) 7.0 (b) 7.30 (c) 8.42
Explanation:
(a) HBr is a strong acid and NaOH is also a very strong base, hence, the solution will be neutral give a pH of 7.0
(b) HClO3
ClO2- + H20 <=> HClO2- + OH-
ClO2- = 0.1*0.08/0.18 = 0.044M
Kb=Kw/Ka=1.0 x 10^-14/1.1 x 10^-2=9.09x10^-13
[OH-] = 2.0 x 10^-7M
p[OH] = 6.7
pH = 7.3
(c) C6H5COOH
C6H5COO- + H20 <=> C6H5COOH + OH-
[C6H5COO-] = 0.1 * 0.80/0.180 = 0.44
Kb=Kw/Ka=1.0 x 10^-14/6.3 x 10-²=1.59x10^-10
[OH] = 1.59x10-10/0.44 = 2.69x10-6
pOH = 5.58
pH = 8.42
The pH of the following compounds in the titration is: (a) HBr - 7;
(b) [tex]\rm HClO_2[/tex] - 7.0275; (c) Benzoic acid - 9.122.
Computation for the pH of the titration reaction
The titration is the neutralization reaction for the acid and base, resulting from the formation of salt.
The pH of the titration reaction can be found as:
[tex]\rm pH=\dfrac{1}{2}\;(pK_a+pK_w+log \;C)[/tex]
The log C value of the reaction of 0.1 M acid with 0.08 M base is 0.044.
The pH with the following compounds is given by:
- HBr
The HBr and NaOH both are strong acids, thus neutralizing each other, and the resultant pH is 7.
- [tex]\rm HClO_2[/tex]
The value of the acid constant is, [tex]\rm pK_a=1.1\;\times\;10^{-2}[/tex]
The value of [tex]\rm pK_w=14[/tex]
The pH for [tex]\rm HClO_2[/tex] is:
[tex]\rm pH=\dfrac{1}{2}\;(14+1.1\;\times\;10^{-2}\;+\;0.044)\\\\ pH=7.0275[/tex]
The pH of the titration reaction of [tex]\rm HClO_2[/tex] is 7.0275.
- Benzoic acid
The value of the acid constant is, [tex]\rm pK_a=4.2[/tex]
The value of [tex]\rm pK_w=14[/tex]
The pH for benzoic acid is:
[tex]\rm pH=\dfrac{1}{2}(14+4.2+0.044)\\\\ pH=9.122[/tex]
The pH of the titration reaction of benzoic acid is 9.122.
Learn more about titration, here:
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