How did this person get the numbers for the reaction tables and what are units f
ID: 921276 • Letter: H
Question
How did this person get the numbers for the reaction tables and what are units for the K formula used to find ph after 23ml of NaOH? could you show work/corrections and is the reaction table number M for molarity or moles or what???
You are titrating 50 ml. of nitrous acid with sodium hydroxide (0.15 M). Equivalence is
reached at 22.0 ml of NaOH delivered. Calculate the pH of the solution in the flask: - before the beginning of the titration - after the delivery of 5 ml. of titrant - at half-equivalence
-at equivalence - after delivery of 23 ml. of titrant.
------------------Give detail Show ALL WORK, UNITS and use ICE/reaction table and ML (no millimoles). Please be clear in how you get numbers for everything like the reaction table... I've asked this before and have gotten incomplete answers.
M * 0.05 0.15* 0.022 M = 0.066 /005 * r/0.05 (0.0033 -0.05 = 7.1 * 10-4 r = 0.000397 pH =-Log(0.000397/0.05) = 2.1 With: 5mL NaOH 0.15M 0.002903...0.000750000397 0.001147 pH = pK, + Login HNO2 0.001147 0.002153PH-2.88 pH-3.15+Log With: 11mL NaOH 0.15M 0.002903..0.00160000397 .001253. pH-3.15+Log With: 22mL NaOH 0.15M 0.002047 0.002047 0.001253H 3.36 0.002903. . . . 0.0033. … …0.000397 0.0.000397.0.0033 0.003697...0.000397 0.00133- r 0.000397 ki = (0.000397 + z)/0.072 * (z)/0-072 = 1.41 * 10 -11 (0.0033- )/0.072 8.44 * 10-12 [OH-] = 0.0003974 8.44 * 10-12 pOH = 2.26 pH = 11.74 With: 23mL NaOH 0.15M = 0.005514 0.072 0.002903...0.00490.000397 pOH = 1.48 pH = 12.52Explanation / Answer
millimmoles of acid = millimoles of base at equivalence point
C x 50 = 0.15 x 22
C = 0.066 M
so concentration of HNO2 = 0.066 M
millimoles of HNO2 = 0.066 x 50 = 3.3
HNO3 Ka = 4.0 x 10^-4
pKa = 3.15
before the beginning of the titration
pH = 1/2 [pKa -logC]
pH = 1/2 [3.15-log 0.066]
pH = 2.16
after the delivery of 5 ml. of titrant
millimoles of NaOH = 5 x 0.15 = 0.75
HNO2 + NaOH -------------------> NaNO2 + H2O
3.3 0.75 0 0 ----------------> initial
2.55 0 0.75 0.75 -----------------> after reaction
pH = pKa + log [NaNO2]/[HNO2]
pH = 3.15 + log (0.75 / 2.55)
pH = 2.62
at half-equivalence
at half equivalence point : pH = pKa
pH = 3.15
at equivalence
at equivalence point only salt NaNO2 remains. its millimoles = 3.3
NaNO2 concentration = millimoles / total volume
= 3.3 / (50 + 22)
= 0.0458 M
strong base weak acid type salt pH formula
pH = 7 + 1/2 [pKa + logC]
pH = 7 + 1/2 [3.15 + log 0.0458]
pH = 7.91
after delivery of 23 ml. of titrant.
millimoles of NaOH = 23 x 0.15 = 3.45
HNO2 + NaOH -----------------> NaNO2 + H2O
3.3 3.45 0 0
0 0.15 3.3 3.3
strong base remains here
NaOH concentraion = 0.15 / (50+23) = 0.00205M
pOH = -log [OH-] = -log (0.00205) = 2.69
pH + pOH = 14
pH = 11.31
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