6. Calculate the value of the K, for the unknown aeid HA from the data at the po

Question

6. Calculate the value of the K, for the unknown aeid HA from the data at the poirt at which 4.70 mL of NaOH have been added to the acid. Read the pH from your graph. This should be clearly marked on your graph in the same manner as mark the pH in 5 above. USE BASE MOLARITY FROM #2. used to 3-75 pH: For each calculation below, show your work. H'loq [A lag Calculate [A 1 produced by the titration reaction. What is [A ] from the acid dissociation reaction? Sum these two [A] values for the [Aleq HAJo Ka

Explanation / Answer

in equilibrium

[H+] = 10^-pH

the pH 3.75

then

[H+] = 10^-3.75 = 0.0001778 M = 1.778*10^-4 M

[A-]eq --> recall htat ratio is

HA <-> H+ + A-

so

[H+] = [A-] = 1.778*10^-4 M

now...

[HA]eq = Molarity initially - dissociated A- = (M - 1.778*10^-4)

For Ka

Ka = [H+]][A-]/[HA]

Ka = (1.778*10^-4)(1.778*10^-4) / ( M - 1.778*10^-4)

Since no molarity is given, assign M = 0.1 M

Ka = (1.778*10^-4)(1.778*10^-4) / ( 0.1 - 1.778*10^-4)

Ka = 3.166*10^-7

Related Questions

Navigate

• Browse (All)
• Browse 6
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.