Determine the pH of a .50-M solution of NH4 OCl. (See Exercise 177.)

Name: Problem 178, Chapter 13: Chemistry An Atoms First Approach
Uploaded: 2023-03-27T12:50:33-08:00
Duration: 8 min 23 s
Channel: Shalini Tyagi
Description: Determine the pH of a .50-M solution of NH4 OCl. (See Exercise 177.)

Determine the pH of a .50-M solution of NH4 OCl. (See...

Key Concepts

Ionic Hydrolysis and pH Calculation

Ionic hydrolysis involves the reaction of salt ions with water, leading to the generation of hydronium or hydroxide ions which influence the pH. Calculating the pH of such a solution typically requires setting up equilibrium expressions based on the hydrolysis reactions and using the associated Ka and Kb values for the conjugate pairs.

Equilibrium Constants (Ka and Kb)

Acid dissociation constants (Ka) and base dissociation constants (Kb) are measures of the strength of acids and bases in water. Using these constants, one can predict the direction and extent of proton transfer in water, which is critical for determining whether a salt solution will be acidic or basic.

Salt Hydrolysis

Salt hydrolysis refers to the chemical reaction that occurs when the ions of a salt react with water to produce an acidic or basic solution. In aqueous media, the ions that are conjugate acids or bases can interact with water by either donating or accepting protons, thereby altering the pH of the solution.

Conjugate Acid-Base Pairs

This concept involves pairs of chemical species that differ by a proton. In the context of a salt solution, one ion is usually the conjugate acid of a weak base, and the other is the conjugate base of a weak acid. The relative strengths of these conjugate pairs determine the net effect on the pH of the solution.

Transcript

00:01 Nh4 ocel is a salt formed by weak base and weak acid and this type of salt h -positive ion is independent of the concentration of salt.

00:11 So i have an h -4 -positive ion in aquas and then it combines with ocl negative.

00:27 This is also in aquas and now i have water.

00:35 In liquid state so this is an equilibrium with h -o -c -l -aquess and nh3 aquas okay so now the degree here this called as the hydrolysis constant okay so the hydrolysis constant this will be given by the concentration of the product divided by concentration of reactant and we have to consider only the aquas ions here so this will be concentration of hocl multiplied by concentration of ammonia okay and divided by we have concentration of nh4 positive and we have concentration of ocl negative okay so now on multiplying and dividing by h positive.

02:05 So now what will happen? we have a new equation.

02:08 Hydrolysis constant, kh is equal to.

02:11 So this will be concentration of h .ocl multiplied by concentration of nh3.

02:19 Then concentration of h positive ion and oh negative ion.

02:29 We will also multiply and divide by oh negative ion here.

02:35 So now dividing by concentration of nh4 positive ion multiplied by ocl negative concentration multiplied by h positive ion concentration multiplied by oh negative ion.

02:54 So kh, so that's equal to kw divided by k a multiplied by k b okay so this is what we will get so now also degree of hydrolysis h that's equal to under root the hydrolysis constant divided by one plus under root of kh again so now here ph so that is given by p -k -a minus log -h now plus log -h okay so now the value for k -a that is given to us for h -o -c -l and this is 3 .5 multiplying ten -raise to minus 8 then value of kb for n h 3 okay so that is given so this is 1 .8 multiplying 10 raised to power minus 5 and k w so that is also given this is 1 .0 multiplying 10 raise to power minus 14 in case it is not given you should remember this value so now putting kh is equal to putting in this equation.

04:50 So this will be 1 .0 multiplying 10 x to power minus 14 divided by 1 .8 multiplying 10 raised to power minus 3, multiplying 3 .5 multiplying 10 raise to power minus 8.

05:09 Okay.

05:10 So now solving this...

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