A student prepared a NaOH solution by adding 5.00 mL of 6.0 M NaOH to a flask an

Question

A student prepared a NaOH solution by adding 5.00 mL of 6.0 M NaOH to a flask and adding distilled water to the 300 mL line. After the solution was stirred, it was poured into a buret; the buret read 0.20 mL. Next, a separate buret was used to dispense 20.10 mL of 0.1046 M HCl into a small flask. Phenolphthalein was added to the acid in the small flask. The acid was then titrated with the prepared NaOH solution. At the endpoint, the NaOH buret read 18.46 mL.

a) How many moles of H+ were used?

b) How many mL of NaOH were used?

c) How many OH- ions were used?

d) What is the concentration of the prepared NaOH solution?

Explanation / Answer

a) 20.10 mL / (1000 mL/L) x 0.1046 mol/L = 0.002102 mol H+

b) 18.46 mL - 0.20 mL = 18.26 mL NaOH

c) I'm assuming you want moles of OH- used... And there are two ways you can calculate this. Theoretically, the moles of H+ = moles of OH-, so the answer could be the same as in part A. Or if you use info from the calculated molarity in part d, you can determine how many moles of OH- you actually used. The numbers are somewhat close.

18.26 mL NaOH / (1000 mL/L) x 0.10 M = 0.001826 mol OH-

(or from part a) 0.002102 mol OH-

d) M1V2=M2V2

M2 = (6.0 M)(5.00 mL)/(300 mL) = 0.10 M

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