Does my math add up?Calculate the moles of NaOH delivered at the equivalence point as determined by the second derivative plot for your careful trial, then enter the values in Table 2.24.178(mL) x 1 / 1000 (mL) x 0.1002(M) = 0.0024709916 mol= 2.47 x 10 ^ -3 molCalculate the number of moles of HCl as experimentally determined by the titration into Table 2.2.47 x 10 ^ -3 molCalculate the concentration of the original HCl solution for then enter the values in Table 2.M = n / VL= 2.47 x 10 ^-3 / 10 x 10 ^-3= 0.247 molar Table 2. Second Derivative Processed Data Trial 2moles of NaOH at equivalence point2.47 x 10 ^ -3moles of HCl2.47 x 10 ^ -3concentration of original HCl solution (M) 0.247

The given calculation for calculating number of mol of NaOH = molarity×volume of solution in L=…

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Chemistry for Today: General, Organic, and Biochemistry

9th Edition

ISBN:9781305960060

Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen

Publisher:Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen

Chapter9: Acids, Bases, And Salts

Section: Chapter Questions

Problem 9.90E

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Does my math add up?

Calculate the moles of NaOH delivered at the equivalence point as determined by the second derivative plot for your careful trial, then enter the values in Table 2.

24.178(mL) x 1 / 1000 (mL) x 0.1002(M) = 0.0024709916 mol

= 2.47 x 10 ^ -3 mol

Calculate the number of moles of HCl as experimentally determined by the titration into Table 2.

2.47 x 10 ^ -3 mol

Calculate the concentration of the original HCl solution for then enter the values in Table 2.

M = n / V_L

= 2.47 x 10 ^-3 / 10 x 10 ^-3

= 0.247 molar

Table 2. Second Derivative Processed Data

	Trial 2
moles of NaOH at equivalence point	2.47 x 10 ^ -3
moles of HCl	2.47 x 10 ^ -3
concentration of original HCl solution (M)	0.247

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