A 0.10mol/l solution of propanoic acid (CH3CH2COOH) is pH 2.96.Calculate the Ka(

Question

A 0.10mol/l solution of propanoic acid (CH3CH2COOH) is pH 2.96.Calculate the Ka(Acid dissociation Constant) and the percentdissociation for this acid.
CH3CH2COOH + H2O <-> CH3CH2COO^- + H3O^+

Explanation / Answer

pH = -log [H+] =2.96 ==> [H+]=10-2.96                                                    = 0.001096 M since the propanoic acid is mono protic  CH3CH2COOH -->CH3CH2COO- + H+                                                          [CH3CH2COO- ]=[H+]=0.001096 M                  Then the concentration of [CH3CH2COOH] =0.1-0.001096 =0.098904M              Ka=[CH3CH2COO-] [H+] / [CH3CH2COOH]            ==>Ka=[ 0.001096 ][ 0.001096 ] / 0.098904                      = 1.21453 X10-7 in 0.1 M acid    0.001096 M dissociated    100 ------------ x there fore percent dissociation =( 0.001096 /0.1) X 100 =1.096 %

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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