A student finds that a volume of 11.51 mL of 0.125 M NaOH is required to neutral

Question

A student finds that a volume of 11.51 mL of 0.125 M NaOH is required to neutralize a 1000 ml sample of a HCl(aq) solution. What is the concentration of HCl in the sample solution. Place the following statements in the correct table column below: Causes [H_3O^+] to increase. Causes [H_3O^+] to decrease. Causes [OH^-] to increase. Causes [OH^-] to decrease. Label the following points on the graph below: a. Pre-equivalence region b. Equivalence point c. Post-equivalence point What volume of 0.181 M HCl is required to neutralize 135. mL of 0.957 M NaOH?

Explanation / Answer

Q1.

NaOH + HCl = H2O + NaCl is the reation

V = 11.51 mL of M = 0.125 M of NaOH

V = 10 mL of HCl, M = ?

apply stoichiometry

1 mol of base = 1 mol of acid

mmol of acid = M*V = M*10

mmol of base = MV = 11.51*0.125 = 1.43875

mmol of acid = mmol of base

M*10 =  1.43875

M =  1.43875/10 = 0.143875 M of Acid

Q2.

when we add ACID:

[H3O+] must increase, and OH- must decrease due to the Kw = [H+][OH-] ratio

when we add Base:

[OH-] must increase, and [H3O+] must decrease due to the Kw = [H+][OH-] ratio

Q3.

pre equivalence region is the region BEFORE, the dratic pH change

equivalence point ---> is the volume point in which the pH changes drastically, approx V = 8 mL

finally post-equivalence point --> must be after the drastic pH, where pH is basic

Q4.

mmol of base = MV = 135*0.957

mmol of HCl = MV = 0.181*V

V = 135*0.957/0.181

V = 717.75 mL

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