Someone PLEASE help me with these 2 questions!!! I\'d GREALTY appreciate it!! Th

Question

Someone PLEASE help me with these 2 questions!!! I'd GREALTY appreciate it!! The answers are provided I just need to see step by step solutions! Thank you sooo much!

Calculate the fluoride ion concentration and pH of a solution containing 0.10 mol of HCl and 0.20 mol of HF in 1 L of solution. Ka for HF equals 6.8 * 10-4.       5

Ans: pH = .994,[F-] = .00133

A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution (a) after 0.020 mol of NaOH is added (neglect any volume change); (b) after 0.020 mol of HCl is added ( again, neglect volume changes).Ka = 1.810-5.        5

a) pH = 4.8, b) pH= 4.68

Explanation / Answer

Calculate the fluoride ion concentration and pH of a solution containing
0.10 mol of HCl and 0.20 mol of HF in 1 L of solution.
Ka for HF equals 6.8 * 10-4.

Answer
Q1.
HF <=> H+ + F-
Initial concentration,   [HF]0 = 0.2, [H+]0 = [HCl]0 = 0.1, [F-]0 = 0
Change in concentration, D[HF] = -x, D[H+] = x, D[F-] = x
Final in concentraion,   [HF]eq = 0.2-x, [H+]eq = 0.1+x, [F-]eq = x
HF <=> H+ + F-
Ka = (0.1+x)*(x)/(0.2-x) = 6.8*10-4

Approximate result:
considering x as small value, so that 0.1+x = x and 0.2-x=0.2, x = 13.6*10-4 = 0.00136
pH = -Log[H+] = -Log(0.1+x) = 1.0

Accurate result:
Ka = (0.1+x)*(x)/(0.2-x) = 6.8*10-4

x*(1+10*x) = 6.8*10-4*(2-10*x),
10*x^2 + x*(1+0.0068) - 13.6*10-4 = 0
x = 0.00133,
[F-] = x = 0.00133
pH = -Loh[H+] = -Log(0.1+0.00133) = 0.994

Q2.
A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00L of solution.
The pH of the buffer is 4.74. Calculate the pH of this solution (a) after 0.020 mol of NaOH is added (neglect any volume change);
(b) after 0.020 mol of HCl is added ( again, neglect volume changes).Ka = 1.8

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