#### Transcript pH and Ka values of Weak Acids

pH and Ka values of Weak Acids AP Chem April 24, 2012 4/13/2015 1 Weak Acids: Calculation of Ka from pH • Will need to use ICE skills for solving equilibrium problems. • Because the concentration of the acid (reactant side) does NOT equal the concentration of the H+ ion (product side) • There is far less than 100% ionization taking place. 4/13/2015 2 A student prepared a 0.10 M solution of formic acid (HCHO2). A pH meter shows the pH = 2.38. a. Calculate Ka for formic acid. b. What percentage of the acid ionized in this 0.10 M solution? 4/13/2015 3 HCHO2 (aq) + H - (aq) + CHO2 (aq) • First, let’s find the [H+] from the pH • [H+] = 10(-2.38) • = 4.2 x 10-3 M • Great, Now for some ICE 4/13/2015 4 HCHO2 (aq) H+ (aq) + CHO2- (aq) I HCHO2 H+ CHO2- 0.10 M 0 0 C E 4/13/2015 5 HCHO2 H+ CHO2- I 0.10 M 0 0 C -4.2 x 10-3 M +4.2 x 10-3 M +4.2 x 10-3 M E 0.10 -4.2 x 10-3 M = 0.0958 4.2 x 10-3 M 4.2 x 10-3 M Assumed from the pH [H+] 4/13/2015 6 So, now for the Ka calculation: H CHO Ka 2 HCHO2 (4.2 103 )(4.2 103 ) Ka = 1.8 x 10-4 0.0958 Is our answer reasonable? Yes, Ka values for weak acids are usually between 10-3 and 10-10. 4/13/2015 7 And what about the percent ionization stuff? • Formula to use: PercentIonization % = 4.2 x 10-3 x 100 0.10 4/13/2015 H equil 100 Acidinitial = 4.2 % 8 Niacin, one of the B vitamins, has the following molecular structure: A 0.020 M solution of niacin has a pH of 3.26. (a) What percentage of the acid is ionized in this solution? (b) What is the aciddissociation constant, Ka, for 4/13/2015 niacin? 9 Niacin Problem #1 • pH = 3.26 [H+] = ? • [H+] = 10-3.26 = 5.50 x 10-4 M • Percent Ionization = [H+]equilibrium x 100 [Acid]Initial = 5.50 x 10-4 M / 0.02 M x 100 = 2.7 % 4/13/2015 10 Solution to Niacin Problem • Ka = [H+] [ niacin ion-] [niacin] Niacin H+ niacin ion I 0.02 0 0 C - 5.50 x 10-4 + 5.50 x 10-4 + 5.50 x 10-4 E 0.02 - 5.50 x 10-4 5.50 x 10-4 5.50 x 10-4 • Ka = (5.50 x 10-4)2 = 1.55 x 10-5 0.019 4/13/2015 11 Using Ka to Calculate pH • Similar to the approach we used in Chapter 15, sometimes using the quadratic equation to solve for the equilibrium concentrations. Once you know the equilibrium concentration of [H+], you can calculate the pH. • Need to have Ka value and the initial concentration of the weak acid • Start by writing equation and equilibrium-constant expression for the reaction. • Let’s calculate the pH of a 0.30 M solution of acetic acid (HC2H3O2) at 250C. 4/13/2015 12 First step: Write the ionization equilibrium for acetic acid: • HC2H3O2(aq) H+ (aq) + C2H3O2- (aq) Second Step: Write the equilibriumconstant expression • Ka = [H+ ] [C2H3O2- ] = 1.8 x 10-5 [HC2H3O2] 4/13/2015 13 Step 3: Set up an ICE calculation I HC2H3O2(aq) H+ (aq) + C2H3O2- (aq) 0.30 M 0 0 C -x M E (0.30 – x) M 4/13/2015 +x M +x M xM xM 14 Fourth Step: Substitute the equilibrium conc into expression. • Ka = [H+ ] [C2H3O2- ] = 1.8 x 10-5 [HC2H3O2] • = (x) (x) = 1.8 x 10-5 (0.30 –x) Solve using quadratic equation: x = 2.3 x 10-3 M • Percent Ionization = [H+]equilibrium x 100 [Acid]Initial % ionization = 0.0023 M x 100 = 0.77% 0.30 M 4/13/2015 15 • Calculate the pH of a 0.20 M solution of HCN (Refer to table 16.2 or Appendix D for the Ka value.) 4/13/2015 16 Solution • HCN (aq) H+ (aq) + CN-(aq) • Ka = [H+ ] [CN-] = 4.9 x 10-10 [HCN] I C E 4/13/2015 0.20 M -x M 0.20 – x 0 +x M xM 0 +x M xM 17 • (x) (x) = 4.9 x 10-10 (0.20 –x) Use quadratic equation to solve for x: x2 = 4.9 x 10-10(0.20 – x) x2 + 4.9 x 10-10x – 9.8 x 10-11 = 0 x = 9.9 x 10-6 = [H+] pH = -log(9.9 x 10-6) pH = 5.00 4/13/2015 18 Second Niacin problem The Ka for niacin is 1.6 x 10-5. What is the pH of a 0.010 M solution of niacin? 1st find the [H+] at equilibrium Initial Change Equilibrium 4/13/2015 Niacin H+ niacin ion 0.010 0 0 -x +x +x 0.010-x x x 19 • Ka = [H+] [niacin ion] = 1.6 x 10-5 [niacin] 1.6 x 10-5 = x2 / (0.010-x) x2 + 1.6 x 10-5 x - 1.6 x 10-7 = 0 x = 3.92 x 10 -4 = [H+] pH = -log(3.92 x 10 –4) pH = 3.41 4/13/2015 20 5.00 mL of 0.250 M HClO3 diluted to 50.0 mL; pH =? 1L 0.250 mol 5.00 mL 0.00125m ol H 1000mL 1 L soln 0.00125mol 0.025M H 0.050L soln pH log(0.025) 1.6 4/13/2015 21 A solution formed by mixing 50.0 mL of 0.020 M HCl with 125 mL of 0.010 M HI. pH=? 1 L soln 0.020 mol 50.0 mL 0.001mol H 1000mL 1 L soln 1 L soln 0.010 mol 125 mL 0.00125mol H 1000mL 1 L soln 0.00225mol H 0.00225mol H 0.0129M H (0.050L 0.125L) 0.175L pH log(0.0129) 1.89 4/13/2015 22