lecture 5

What is the pH when 25.0 mL of 0.200 M of CH3COOH has been titrated with 35.0 mL of 0.100 M NaOH?

1) Determine moles of acetic acid and NaOH before mixing: CH3COOH: (0.200 mol/L) (0.0250 L) = 0.00500 mol NaOH: (0.100 mol/L) (0.0350 L) = 0.0035 mol 2) Determine moles of acetic acid and sodium acetate after mixing: CH3COOH: 0.00500 mol - 0.00350 mol = 0.00150 mol CH3COONa: 0.0035 mol 3) Use the Henderson-Hasselbalch Equation: pH = 4.752 + log [(0.00350 mol/0.060 L) / (0.0015 mol/0.060 L)]pH = 4.752 + log 2.333 pH = 4.752 + 0.368 = 5.120 Note the inclusion of the volumes. They simply cancel out, so most HH problems can be solved with moles rather than molarities.

The pH of "Solution 1"is 3.5 and the pH of "Solution 2" is 3.8. How many times more acidic is solution 1 than solution 2?

The hydrogen ion concentration of "Solution 1" is calculated to be, [H+ ]= 10−3.5 ≈3.2 × 10−4 M, and the hydrogen ion concentration of "Solution 2" is calculated to be, [H+] = 10^2−3.8 ≈ 1.6 × 10−4 M. Thus, "Solution 1" is approximately twice as acidic as "Solution 2"since,

What is the hydrogen ion concentration of a solution labeled "Solution A", whose pH is 5.60?

Using the equation, pH = − log [H+] , we can solve for [H+] as, − pH = log [H+] , [H+] = 10−pH The hydrogen ion concentration of "Solution A" is, [H+] = 10^−5.6 ≈ 2.51 × 10−6 M. Always check our solution by computing −log (2.5 × 10−6) ≈ 5.6.

Aspirin has a pKa of 3.4. What is the ratio of A¯ to HA in: (a) the blood (pH = 7.4) (b) the stomach (pH = 1.4)

pH = pKa + log ([A¯] / [HA]) 7.4 = 3.4 + log ([A¯] / [HA]) 7.4 - 3.4 = 4 = log ([A¯] / [HA]) 104 = 10000 = [A¯] / [HA] 1.4 = 3.4 + log ([A¯] / [HA]) 1.4 - 3.4 = - 2.0 = log ([A¯] / [HA]) 10-2 = 0.01 = [A¯] / [HA]

Calculate the concentration of hydrogen ions in the following acetic acid solution. pKa= 4.76, ratio 0.5 [HA]/[A]

pH = pKa + log[A]/[HA] pH= 4.76 + log(2)= 5.06

A buffer is made from acetic acid (HC2H3O2) and sodium acetate (NaC2H3O2). 1.5 moles of acetic acid and 2.5 moles of sodium acetate are added to enough water to make 1.5 liter of solution. The ionization constant (Ka) for acetic acid is 1.8 x 10-5 .

pH = pKa + log[HA]/[A] pH=-log(1.8 x 10^-5) + log [(2.5mol/1.5L)/(1.5mol/1.5L)]=4.97

What is the pH of a mystery solution with hydrogen ion concentration, 0.050 M?

pH = − log [H+ ]= − log [0.05] ≈ 1.30.

What is the pH of lemon juice that has a hydrogen ion concentration of 5.4 x 10-3 M?

pH = −log [H+] = − log [5.4 × 10−3] ≈ 2.27 or 2.3.

what is the pH of a buffer solution containing 0.9M HC2H3O2 and 1.10M NaC2H3O2?

pH= pka + log [A-]/[HA] 4.83