Consider a .10M solution of HNO2 has a Ka=4.5*10^4. A researcher was tasked to m
ID: 1004890 • Letter: C
Question
Consider a .10M solution of HNO2 has a Ka=4.5*10^4. A researcher was tasked to make a 100mL solution buffer by titrating with .10M NaOH
a) what is the initial PH Of solution?
B)the researcher misread the instructions and erroneously added 50mL of NaOH solution to 50mL of HNO2 solution. what is the PH of that solution?
C)The researcher read the instruction but still did not know what to do and therefore decided instead to simply add .500g of NaNO2 to 100mL of HNO3 solution AND WISH FOR LUCK(what is the pH of this solution)
d) HELP THE RESEARCHER and determine what the actual volumes of HNO3 and NaOH were needed to make 100mL of this buffer?
Explanation / Answer
Solution:
a) by using hendersen-hasselbalch equation the pH are computed using given values,
pH = pKa + log {[salt]/[acid]}
pH = log (4.5 E04) + log {0.1/0.1}
pH = 4.65
b) moles of HNO2 = 0.05 * 0.1 = 0.005 moles
moles of NaOH = 0.05 * 0.1 = 0.005 moles
Total volume = 50 + 50 = 100 ml
[HNO2] = 0.005 / 0.1 = 0.05 M
[NaOH] = 0.005 / 0.1 = 0.05 M
by using hendersen-hasselbalch equation the pH are computed using given values,
pH = pKa + log {[salt]/[acid]}
pH = log (4.5 E04) + log {0.05/0.05}
pH = 4.65
no change observed from initial pH as equal amount is added.
c) moles of NaNO2 = 0.5 / 68.99 = 0.007 moles
moles of HNO3 = 0.1 M
Total volume = 100 mL
[NaNO2] = 0.007 / 0.1 = 0.07 M
by using hendersen-hasselbalch equation the pH are computed using given values,
pH = pKa + log {[salt]/[acid]}
pH = log (4.5 E04) + log {0.07/0.1}
pH = 4.49
d) 0.1 M of HNO3 needs 10 ml of 1M acid and 40 g of NaOH in 90 ml
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