1)
HCN dissociates as:
HCN -----> H+ + CN-
3*10^-3 0 0
3*10^-3-x x x
Ka = [H+][CN-]/[HCN]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((6.2*10^-10)*3*10^-3) = 1.364*10^-6
since c is much greater than x, our assumption is correct
so, x = 1.364*10^-6 M
So, [H+] = x = 1.364*10^-6 M
use:
pH = -log [H+]
= -log (1.364*10^-6)
= 5.8652
Answer: 5.87
Only 1 question at a time please
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Q1: Determine the pH solution contains 1.0 M of HCN (ka = 6.2 x 10-10) and HNO2 (ka = 4.0 x 10-4)? Correct answer is pH = 1.35
Given that Ka for HCN is 6.2*10^-10, calculate the pH of a 0.15 M KCN solution.
Calculate the pH of a solution that is 1.50 M in HCN (Ka = 6.2 x10-10) and 0.50 M in NaCN. A. 3.44 B. 8.73 C. 9.68 D. 9.21 E. 3.63
Calculate the following: a. The pH of a 500.0 mL buffer solution containing 0.75 M HCN (Ka = 6.2 x 10^-10) and 0.55 M NaCN b. The pH of the above buffer after the addition of 100.0 mL of 1.0 M NaOH. c. The pH of the buffer if 100.0 mL of 1.0 M HCl was added to the solution in part a.
Solve for the pH of a 0.0986 M solution of NaCN. Given Ka (HCN)= 6.2 * 10-10 Please have two decimal places for your answer.
Calculate the pH of a solution that contains 3.25 M HCN (Ka = 6.2 × 10–10), 1.00 M NaOH and 1.50 MNaCN. Question 14 options: A) 8.28 B) 7.46 C) 9.25 D) 8.86 E) none of these
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A solution contains 0.50 M hydrocyanic acid (HCN; Ka = 6.2 × 10–10 at 25 °C) and 0.25 M sodium cyanide (NaCN) at 25 °C. Calculate the pH of this solution. Show (or explain) your calculation.