Here are all my calculations from the lab Starting out pH 2.78 Burrette Volume 5
ID: 1019112 • Letter: H
Question
Here are all my calculations from the lab
Starting out pH 2.78
Burrette Volume 50.00mL Dispensed Volume 0.00mL Flask 25.10
Volume of Burrette 49.00 DV 1 Flask 26.10 pH 4.28
VB 48.000 mL DV 2mL Flask 27.09 pH 4.76
VB 47.000 mL DV 3 mL Flask 28.09 pH 5.23
VB 46.00 mL DV 4 mL Flask 29.08 pH 8.95 PINK!!!!!!!
VB 45.00 mL DV 5 mL flask 30.08 Ph 12.52mL
VB 44.00mL DV6mL flask 31.08 pH 12.80
VB 43.00 mL DV 7 mL flask 32.08 pH 12.96
VB 42.00mL DV 8mL flask 33.08 Ph13.07
VB 41 DV 9 mL flask 34.08 pH 13.15
I am sooo wrong on all of this... I think I need some major help!!!!
2.Find the equivalence point on the graph. What is the equivalence volume of NaOH at this point?
HERE IS WHERE THE HELP NEEDED STARTS! JUST IGNORE MY ANSWERS AND REWORK THEM PLEASE!!!!
3. a. Calculate the unknown molarity of the diluted acetic acid from the volumes of acid and base at the equivalence point and the molarity of the NaOH Ma × Va = Mb × Vb.
Volume of NaOH: (Vb)=4.00mL
Molarity of NaOH: (Mb)=1.0 M
Volume of acid, (Va)=25.1
Molarity of acid (Ma)=x
1.0M*4mL/25.1 mL=.1594
Molarity of acid
=.16 M
b. Once you find the molarity of your diluted solution use that to calculate the molarity of the original solution using the equation M1 × V1 = M2 × V2 a second time.
Initial Volume of acid(V1)=mL
Initial Molarity of acid (M1)=x
Final Volume of acid (V2)= 29.08 mL
Final Molarity of acid (M2)=.16
M1V1=M2V2
m1=M2V2/V1
=.1594M *29.08 mL/5.0mL=
Inital Molarity
=.93M
The equivelance point is 4 with a pH of 8.95. Titration of Acetic acid Volume of Acetic Acid mL show trendline Slope: 1.677 Intercept: 2.166 SAVE GRAPHExplanation / Answer
2. Equivalence volume on the graph = 4 ml ; pH = 8.95
3. molarity of diluted acetic acid solution = 1 M x 4 ml/(25.1 + 4)ml = 0.14 M
4. molarity of original acetic acid (unknown) = 0.14 M x 29.1 ml/25.1 ml = 0.16 M
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