an A: Standardization of NaOH solution Trial #1 Trial #2 Trial #3 Trial #4 Mass
ID: 547769 • Letter: A
Question
an A: Standardization of NaOH solution Trial #1 Trial #2 Trial #3 Trial #4 Mass KHCsH404, grams .504 inal buret reading, ml nitial buret reading, mL S Volume NOH used, mL 12e.1m,LINN Part B: Determination of the molar mass f an uninown acid 1 Trial #1 | Trial #2 | Trial #3 Trial #4 nknown sample umber: Mass unknown acid 209 2309 8534 Final buret reading, mL | 32.5%43aq1dBysst Initial buret reading,m 4248 Volume NaOH usem1T27 23 l Part C: Analysis of Acetic Acid in Vinegar Trial #4 Trial #1 | Trial #2 | Trial #3 Volume of vinegar, mL S.con s.ooud Final buret reading, mL 43.304Dame 31oS seo Yo 9amalthm gsulls and Averagevolume tial buret reading, mL Rev 8-23-17Explanation / Answer
Part A
Trial #1
Trial #2
Trial #3
Trial #4
Mass KHC8H4O4 (g)
0.504
0.502
0.501
0.500
Mole(s) KHC8H4O4 (mole) [Molar mass of KHC8H4O4 = 204.22 g/mol]
(0.504)/(204.22) = 0.002468
(0.502)/(204.22) = 0.002458
(0.501)/(204.22) = 0.002453
(0.500)/(204.22) = 0.002448
Volume of NaOH used (mL)
20.72
19.44
22.89
19.86
Molarity of NaOH used (M)
0.1191 (see calculation below)
0.1264
0.1071
0.1233
Average molarity of NaOH used (M)
¼*(0.1191 + 0.1264 + 0.1071 + 0.1233) = 0.118975
Sample calculation:
NaOH reacts with KHC8H4O4 as below.
NaOH(aq) + KHC8H4O4(aq) ----------> KNaC8H4O4(aq) + H2O(l)
As per the stoichiometric equation,
1 mole NaOH = 1 mole KHC8H4O4.
Therefore, 0.002468 mole KHC8H4O4 = 0.002468 mole NaOH
Molarity of NaOH = (moles of NaOH)/(volume of NaOH in L) = (0.002468 mole)/[(20.72 mL)*(1 L/1000 mL)] = 0.1191 mol/L = 0.1191 M.
Part B
Trial #1
Trial #2
Trial #3
Mass unknown acid, g
0.250
0.250
0.253
Volume NaOH used, mL
27.71
27.61
28.61
Mole(s) NaOH used, mL
0.003297 (see calculation below)
0.003285
0.003404
Mole(s) acid
0.003297
0.003285
0.003404
Molar mass of unknown acid (g/mol)
75.8265 (see calculation below)
76.1035
74.3243
Average molar mass of unknown acid (g/mol)
1/3*(75.8265 + 76.1035 + 74.3243) = 75.4181 75.418
Sample calculation:
Mole(s) of NaOH used = (volume of NaOH in L)*(average molarity of NaOH) = (27.71 mL)*(1 L/1000 mL)*(0.118975 M)*(1 mol/L/1 M) = 0.003297 mole.
Let the unknown acid be denoted as HA. Write down the neutralization reaction.
HA(aq) + NaOH(aq) --------> NaA(aq) + H2O(l)
As per the stoichiometric equation,
1 mole HA = 1 mole NaOH.
Therefore, 0.003297 mole NaOH = 0.003297 mole HA.
Moles of HA = mass of HA taken/(molar mass of HA)
Therefore,
0.003297 mole HA = (0.250 g)/(molar mass of HA)
====> molar mass of HA = (0.250 g)/(0.003297 mole) = 72.8265 g/mol
Trial #1
Trial #2
Trial #3
Trial #4
Mass KHC8H4O4 (g)
0.504
0.502
0.501
0.500
Mole(s) KHC8H4O4 (mole) [Molar mass of KHC8H4O4 = 204.22 g/mol]
(0.504)/(204.22) = 0.002468
(0.502)/(204.22) = 0.002458
(0.501)/(204.22) = 0.002453
(0.500)/(204.22) = 0.002448
Volume of NaOH used (mL)
20.72
19.44
22.89
19.86
Molarity of NaOH used (M)
0.1191 (see calculation below)
0.1264
0.1071
0.1233
Average molarity of NaOH used (M)
¼*(0.1191 + 0.1264 + 0.1071 + 0.1233) = 0.118975
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