
.20 M NaOH is the Molarity
Part 1 Data is .20 M NaOH
Hi my apologies to say that part b & part c requires additional data points A, B, C, and D from part "a". Hence i am gonna answer part a at this moment.
NaOH + HCl ==> NaCl + H2O
1 mol NaOH reacts with 1 mol HCl
M1V1 = M2V2 formula we can use here
M1 = 0.100 M HCl, V1 = 20.00 mL
M2 = 0.20 M NaOH, V2 = to be find out
V2 = (M1V1 / M2) = (0.100 M x 20.00 mL) / 0.20 M = 10 mL
Answer: Volume of NaOH needed to reach the equivalence point = 10.00 mL
Hope this helped you!
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1) A student titrated 20.0 mL of 0.410 M HCl with 0.320 M NaOH and collected the following data: Number V of NaOH solution added, mL PH # V of NaOH added, mL PH 1 0.00 .39 12 22.00 1.56 2 2.00 .46 13 24.00 1.93 3 4.00 .54 14 24.50 2.09 4 6.00 .62 15 25.00 2.35 5 8.00 .70 16 25.50 3.06 6 10.00 .78 17 26.00 11.40 7 12.00 .87 18 26.50 11.80 8 14.00 .96 19...
Data:
standaed HCL concentration : 0.1109 M
Accepted NaOH molarity: 0.0979 M
CALCULATIONS:
NOTE: Molarity (M) is #mole of solute in one L of solution
(units: mole solute/L solution)
HCl(aq) + NaOH(aq) ------------> NaCl(aq) + H2O (1)
1)Using the equation on the first page of this experiment solve
it for MNaOH (molarity of the now standardized base).
2)Calculate MNaOH for each run.
3)Find the average molarity of this NaOH solution.
4) Determine the % error.
Calculate the [HCL], [H3O+], pH, pOH and concentration of
[OH-].
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Calculate and enter the molarity of your
three HCl standardization trials using the volume of standardized
NaOH solution required for each and the average molarity of the
NaOH solution from the standardization trials with KHP. You should
report 3 significant figures, e.g. 0.488 M.
I need to find the average molarity of my NaOH to solve this,
but I am unsure on how to solve for it. It had a concentration of
0.1 M. Thank you!
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Questions:
A. From the standardization data, calculate the molarity of the
sodium hydroxide solution for each trail. Average the values and
enter the average in the Standardization Data Table.
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