TUe) Acias and Bases (1a) Calculate the pH of a 0.20 M HNO2/0.20 M NaNO2 buffer.
ID: 590360 • Letter: T
Question
TUe) Acias and Bases (1a) Calculate the pH of a 0.20 M HNO2/0.20 M NaNO2 buffer. What is the p after the addition of 10.0 mL of 0.10 M NaOH to 90.0 mL of the buffer? Ka (HNO2) H of the solution = 4.0×10-4 cular experiment, Candy Chemist, our favorite acid-base buffer savant, must a buffer with a pH = 8.50. In the laboratory, she finds 1.00 M acetic acid (K.- ) and 1.00 M ammonia (Ks = 1.76×10-5). Which reagent should be usedto prepare (1b) prepare the buffer? CH,COOH should bt vse dto In addition to the above question, your assignment is to calculate the mole ratio of acid to conjugate base (or base to conjugate acid), which can be used to determine the best reagent to prepare the buffer. Is the buffer better for additions of acid or base? Explain.Explanation / Answer
1a) the buffer is HNO2 and NaNO2
[HNO2] = 0.20 M and [NaNO2] =0.20 M
Ka of HNO2 = 4.0x10-4
and pKa = 3.3979
The pH of buffer is calculated using Hendersen equation
pH = pKa + log [conjugate base]/[acid]
= 3.3979 + log 0.2/0.2
= 3.3979
Now 10mL of 0.1M NaOH is added to 90mL of buffer
HNO2 + Na OH ---------------------> NaNO2 + H2O
90x0.2 0 90x0.2 - initial mmoles
- 10x0.1 - - change
17 0 19 - after
Now the pH of solution i
pH = 3.3979 + log 19/17
= 3.4462
1b)pKa of acetic acid = 4.7544 and pKb of ammonia is 4.7544
If the pH of buffer required is 8.5 , the chemist choses ammoni buffer as
pH of ammonia buffer is
pH = 14-pOH
= 14-4.7544
=9.2455 if [NH3] = [NH4+] in the buffer
Thus by adjusting the ratio of NH4+ and NH3 she can easily make the buffer of 8.5
ii) We know
pH = pKa + log [conjugate base]/[acid]
or pOh = pKb + log [conjugate acid]/[base]
The required pH = 8.5 , then pOH = 5.5
Thus
5.5 = 4.7544 + log [conjugate acid]/[base]
[conjugate acid]/[base] = 5.56
The buffer will be better for addition of base as the [conjugate acid] is 5.56 times higher than the base
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.