A mixture is prepared using 30.00 mL of a 0.077 M weak base, B (rho Ko_(HB+) = 4
ID: 901497 • Letter: A
Question
A mixture is prepared using 30.00 mL of a 0.077 M weak base, B (rho Ko_(HB+) = 4.573) and 30.00 mL of a 0.025 M hydrochloric acid, HCI. 1.00 ml of 1.75 Times 10^-4 M of a chromophoric indicator is added to the mixture, and it is further diluted to 100.0 ml with deionized water. The absorbance of the diluted mixture measured at 550 nm in a 5.0 cm cell was 0.155. The molar absorptivity (epsilon) values for Hin and its deprotonated form In^- at 550 nm are 2.45 Times 10^4 M^-1cm^-1 and 1.43 Times 10^4 M^-1cm^-1 respectively. What is the pH of the diluted mixture? What are the concentrations of the protonated and deprotonated form of the indicator found in the diluted mixture? Determine the numerical value of pKa of the indicator, pK_a Show all calculations.Explanation / Answer
a.
0.077 mol/L x 0.030 L = 0.00231 mol B (initial state)
0.025 x 0.030 L = 0.00075 mol HCl used to neutralize B
[BH+] = 0.00075 mol/0.060 L = …
[B] = 0.00150 mol/0.060 L =…
pH = pKa + log([B]/[ BH+]) =
= 4.573+ log(0.00150/0.00075) = 4.874
b.
0.155 = 2.45x104· CHIn· 5 + 1.43x104· CIn-· 5
CHIn + CIn- = 1.75·10-4 mol/L x 1mL/100mL = 1.75·10-6 M
Solve the system of these 2 equations.
Write
CHIn = m · 10-6 M
CIn- = (1.75-m) · 10-6 M
0.155 = 2.45x104· CHIn· 5 + 1.43x104· CIn-· 5
0.031 = 2.45x104· m · 10-6 + 1.43x104· (1.75-m) · 10-6
3.1 = 2.45m + 1.43· (1.75-m)
0.597 = 1.02 m
m = 0.585
CHIn = 0.585 · 10-6 M
CIn- = 1.165· 10-6 M
d.
[H+] = 10-pH = 10-4.874 = 1.337x 10-5 M
Ka = [H+][In-]/[HIn]
= 1.337x 10-5 x 1.165· 10-6 / 0.585 · 10-6 = 2.66x10-5
pKa = -logKa = 4.57
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.