Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Consider the titration of a 24.6 mL sample of 0.120 M RbOH with 0.105 M HCl. Det

ID: 1040991 • Letter: C

Question

Consider the titration of a 24.6 mL sample of 0.120 M RbOH with 0.105 M HCl. Determine each of the following.
A) the initial pH B) the volume of added acid requires to reach the equivalence point C) the pH at 5.8 mL of added acid D) the pH at the equivalence point E) the pH after adding 5.9 mL of acid beyond the equivalence point Consider the titration of a 24.6 mL sample of 0.120 M RbOH with 0.105 M HCl. Determine each of the following.
A) the initial pH B) the volume of added acid requires to reach the equivalence point C) the pH at 5.8 mL of added acid D) the pH at the equivalence point E) the pH after adding 5.9 mL of acid beyond the equivalence point
A) the initial pH B) the volume of added acid requires to reach the equivalence point C) the pH at 5.8 mL of added acid D) the pH at the equivalence point E) the pH after adding 5.9 mL of acid beyond the equivalence point

Explanation / Answer

1)

The reaction happening will be

HCl + RbOH ----- RbCl + H2O

pOH = -log(OH-) = -log(0.120) = 0.9208

pH = 14 - pOH = 14 - 0.9208 = 13.0792

2)

At equivalence point

number of moles of acid = number of moles of base

V/1000 * 0.105 = 24.6/1000 * 0.120

V = 28.114 mL

3)

Number of moles of acid = 5.8/1000 * 0.105 = 0.000609 moles

Number of moles of base = 24.6/1000 * 0.120 = 0.002952 moles

Number of moles of base left = 0.002952 - 0.000609 = 0.002343 moles

Volume of solution = 24.6 + 5.8 = 30.4 mL

Molarity of OH- = 0.002343/30.4 * 1000 = 0.07707

pOH = -log(0.07707) = 1.113

pH = 14 - pOH = 14 - 1.113 = 12.887

4)

pH at equivalence point will be 7, since the titration of strong acid and strong base

5)

Number of moles of acid = 34.014/1000 * 0.105 = 0.00357147 moles

Number of moles of base = 24.6/1000 * 0.120 = 0.002952 moles

Number of moles of acid left = 0.00357147 - 0.002952 = 0.00061947 moles

Volume of solution = 24.6 + 34.014 = 58.614 mL

Molarity of H+ = 0.00061947/58.614 * 1000 = 0.01056863547

pH = -log(0.01056863547) = 1.975