0.015 M CH2FCOOH (Ka CH2FCOOH = 2.6 x10-3), the change in CH2FCOOH can not be neglected because.  a. Acid Dissociation constant, Ka, has a negative exponent. b. pH is not within +/- pka unit of pka of CH2FCOOH c. initial CH2FCOOH concentration is less than 0.1 M

Organic Chemistry
9th Edition
ISBN:9781305080485
Author:John E. McMurry
Publisher:John E. McMurry
Chapter20: Carboxylic Acids And Nitriles
Section20.3: Biological Acids And The Henderson–hasselbalch Equation
Problem 5P: Calculate the percentages of dissociated and undissociated forms present in the following solutions:...
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In calculating pH by ICE method of 0.015 M CH2FCOOH (Ka CH2FCOOH = 2.6 x10-3), the change in CH2FCOOH can not be neglected because. 

a. Acid Dissociation constant, Ka, has a negative exponent.

b. pH is not within +/- pka unit of pka of CH2FCOOH

c. initial CH2FCOOH concentration is less than 0.1 M

d. dissociation of [CH2FCOOH] will be more than 5% of initial [CH2FCOOH]

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To protect corrosion of materials made of iron, the surface of the iron material is electroplated with another metal. Given below information, which metal will be best to “galvanize” the material ?

O2  + H2O + 2e- ->   4OH-(aq) .... E0 = 0.401 v

 Fe2+(g) + 2e- - ->   Fe ............ E0 =(-) 0.440 v

Cu2+(g) + 2e- - ->   Cu ...........E0 = 0.337 v

Ni2+(g) + 2e- - ->   Ni ............ E0 = (-)0.250 v

Zn2+(g) + 2e- - ->   Zn ............ E0 =(-) 0.763 v

 

a. Zn

b. Ni

c. Cu

d. Fe2+

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Given :  C (s) + H2O(g) <-> CO(g) + H2(g)      

Which of the following will shift equilibria to form more CO gas ?

a. Addition of H2O vapour..

b. Addition of H2(g).

c. Decreasing the volume of reaction vessel.

d. Addition of C (s).

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 An enzyme will have following effect on a reaction except

 

a. It lowers the amount of energy needed to initiate the reaction.

b. It participates in the reaction.

c. It changes the Equilibrium constant of the reaction.

d. It immobilizes reactants of the reaction.

--------------Which of the following is not going to make a buffer ?

a. Dissolving 2g of sodium dihydrogen phosphate in water

b. Dissolving of solid 0.2 g sodium hydroxide in Hydrochloric acid solution.

c. Addition of a few drops of HCl in 0.1 M Naacetate

d. Mixing of equal volumes of formic acid and sodium formate solution of same Molarities.

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