Solveeit Logo
Login

Question

Question: Find out ka for acid when degree of hydrolysis of \( 0.1M{{ }}C{H_3}COONa \) is \( 1\% \) : - (A) ...

Find out ka for acid when degree of hydrolysis of 0.1MCH3COONa0.1M{{ }}C{H_3}COONa is 1%1\% : -
(A) 10−510 - 5
(B) 10−910 - 9
(C) 10−710 - 7
(D) 10−1310 - 13

Explanation

Solution

A chemical reaction accomplishes chemical equilibrium when the pace of forward reaction and that of the converse reaction is the same. Additionally, since the rates are equivalent and there is no net change in the convergences of the reactants and the items.

Complete Step By Step Solution:
An answer of sodium acetic acid derivation (an essential salt of acidic corrosive) and acidic corrosive can go about as a cushion to keep a moderately consistent pH level. This is helpful particularly in biochemical applications where responses are pH-subordinate in a somewhat acidic reach (pH4−6)\left( {pH{{ }}4-6} \right) . Sodium acetic acid derivation is utilized as the carbon hotspot for refined microscopic organisms. Sodium acetic acid derivation is likewise helpful for expanding yields of DNA confinement by ethanol precipitation. Sodium acetic acid derivation might be added to food as a flavoring, in some cases as sodium diacetate, a balanced complex of sodium acetic acid derivation and acidic acid. It is frequently used to give potato chips as salt and vinegar flavor.
Balance reaction is
CH3COOH−+H2O⇌CH3COOH+OH−C{H_3}COO{H^ - } + H2O \rightleftharpoons C{H_3}COOH + O{H^ - }
0.1×(1−h)0.1×h0.1×h0.1 \times \left( {1 - h} \right){{ }}0.1 \times h0.1 \times h
Kh=[CH3COO][OH−][CH3COOH]=(0.1×h)(0.1×h)0.1(1−h){K_h} = \dfrac{{\left[ {C{H_3}COO} \right]\left[ {O{H^ - }} \right]}}{{\left[ {C{H_3}COOH} \right]}} = \dfrac{{\left( {0.1 \times h} \right)\left( {0.1 \times h} \right)}}{{0.1\left( {1 - h} \right)}}
hh is little (1−h→1)\left( {1 - h \to 1} \right)
5.6×10−10=0.1×h25.6 \times 10 - 10 = 0.1 \times {h_2}
or on the other hand
h2=5.6×10−10/0.1{h_2}{{ }} = {{ }}5.6 \times 10 - 10/0.1 =56×10−10c= 56 \times 10 - 10c
Level of hydrolysis is h=7.48×10−5h = 7.48 \times 10 - 5
[OH−]=Ch=0.1×7.48×10−5=7.48×10−6M\left[ {O{H^ - }} \right] = Ch = 0.1 \times 7.48 \times 10 - 5{{ }} = 7.48 \times 10 - 6M
[H+]=Kw/[OH−]=10−14/7.48×10−6=1.33×10−9M\left[ {{H^ + }} \right] = {{ }}{K_w}/\left[ {O{H^ - }} \right]{{ }} = {{ }}10 - 14/7.48 \times 10 - 6{{ }} = 1.33 \times 10 - 9M
pH=−log⁡[H+]=−log⁡(1.33×10−9)=8.88pH = - \log \left[ {{H^ + }} \right] = - \log \left( {1.33 \times 10 - 9} \right) = 8.88
So the answer option B.

Additional Information:
Applications:
Sodium acetic acid derivation is utilized in the material business to kill sulfuric corrosive waste streams and furthermore as a photoresist while utilizing aniline colors. Sodium acetic acid derivation is utilized to alleviate water harm to concrete by going about as a solid sealant, while likewise being naturally amiable and less expensive than the ordinarily utilized epoxy elective for fixing concrete against water permeation.

Note:
For research facility use, sodium acetic acid derivation is cheap and generally bought as opposed to being orchestrated. It is once in a while created in a research facility try by the response of acidic corrosive, ordinarily in the 5−8%5-8\% arrangement known as vinegar, with sodium carbonate ("washing pop"), sodium bicarbonate ("preparing pop"), or sodium hydroxide. Any of these responses produce sodium acetic acid derivation and water.