calculate the pH of 25.00 mL of 0.250 M HNO2 as it is being titrated with 0.100 M NaOH.
2007-04-11 13:31:49 · 2 answers · asked by mheg 1 in Science & Mathematics ➔ Chemistry
One point for every 1 mL of NaOH
2007-04-11 13:51:18 · update #1
Ka = 4.5 10^-4 pK= 3.35
a) initial solution 25 mL of 0.250 M HN02
(HNO2)= 0.250 -x (H+)=x (NO2-)= x
4.5 10^-4 = (x)(x) / 0.250-x
x =0.0106 M pH=1.97
b) after adding 5.0 mL of 0.100 M NaOH
moles HNO2 = 0.250 M ( 0.0250 L )= 0.00625
moles NaOH = 0.0050 L ( 0.100 M) = 0.00050 mole OH-
0.00625 mole HNO2 + 0.00050 mole OH- >> 0.00575 mole HNO2 and 0.00050 mole NO2-
Total volume = 0.0300 L
Initial concentration HNO2 = 0.00575 / 0.0300 = 0.192 M
Initial concentration NO2- = 0.00050 / 0.0300= 0.0167 M
pH = pK + log (NO2-) /(HNO2)= 3.35 + log 0.0167/0.192= 2.29
you can calculate for other adding of NaOH
2007-04-12 05:19:20 · answer #1 · answered by Anonymous · 0⤊ 0⤋
How about a Ka for HONO. Also, what volume intervals do you want for NaOH?
2007-04-11 20:39:10 · answer #2 · answered by cattbarf 7 · 0⤊ 0⤋