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Dhamnekar Winod
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pH determination problems

Post by Dhamnekar Winod »

Determine whether each of the following 1 M solutions is acidic, basic, or neutral.
a.NaHS
b.NaBr
c.NH4Ac (remember Ac = acetate)
d.KNO2
e.KHSO3
f.NaH2PO4

I am working on these questions.
Last edited by Dhamnekar Winod on Sat Apr 03, 2021 7:41 am, edited 3 times in total.
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Re: pH determination problems

Post by ChenBeier »

What are strong acid, base and weak acid, base. What are the corresponding species in every example?
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Re: pH determination problems

Post by Dhamnekar Winod »

All the 1 M aqueous solutions of the substances in the questions a, b, c, d, e, f are acidic.
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
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Re: pH determination problems

Post by ChenBeier »

No
A. Alkaline, B. Neutral C. Neutral, D. Alkaline, E. Acidic, F. Acidic

Explain why.: in A compare NaOH and H2S
The same in the other questions
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Re: pH determination problems

Post by Dhamnekar Winod »

You are correct. Answer for question a is alkaline. Following is my chemistry work.

I computed the pH of 1.0 M solution of NaHS as follows:- NaHS + H2O ⇌ NaOH + H2S

\(K_a of HS = 1.2e-13 = \frac{[NaOH])[H_2S]}{[1.0M]}\) Concentration of H2S is =\(\sqrt{1.2e-13}=[3.46e-7] \) We got concentration of H2S. Ionization of H2S is

H2S ⇌ H⁺ + HS

\(K_a of H_2S = 8.9e-8 = \frac {[H^+][1.0M ]}{[3.46e-7]}=3.08e-13\) Hence pH of 1.0 M solution of NaHS=-log(3.08e-13)=12.51.

Hence the solution is alkaline.

Is my computation of pH correct?
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
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Re: pH determination problems

Post by ChenBeier »

Yes correct
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Re: pH determination problems

Post by Dhamnekar Winod »

Your answer to question e for pH of 1 M solution of KHSO3(Potassium Hydrogen Sulfite) is acidic.
But my pH computation work result is different.

KHSO3 + H2O ⇌ H2SO3 + KOH

\(K_a = \frac{K_w}{K_b}= \frac{1e-14}{1.5e-2}=\frac{x\cdot x}{1 M}= 6.66e-13,x= 8.16e-7,pOH=6.088,pH=7.911 \) So, the 1 M solution of KHSO3 is alkaline.

What is wrong with this pH computation?
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
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Re: pH determination problems

Post by ChenBeier »

Used wrong formula H2SO3 is a middle strong acid. pKa1 is 1.8 pka2 is 7.

pH =( pk1 + pk2)/2

pH =(1.8 +7)/2 = 4.4
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