Question

1. What is the pH of 1.85 g of HCN (MM = 27.02: Ka = 6.2 * 10-19) dissolved in 3.50 L of water? A. 1.165 B. 1.708 C.5.186 D. 5.458 E. 8.814

please give an explanation as to how the answers were derived including explanation within the steps. I am stuck once I get my Ka=[x][x]/(0.0195-x) and am unsure how to progress from here. Thank you!

Answer 1

Molar mass of HCN,
MM = 1*MM(H) + 1*MM(C) + 1*MM(N)
= 1*1.008 + 1*12.01 + 1*14.01
= 27.028 g/mol


mass(HCN)= 1.85 g

use:
number of mol of HCN,
n = mass of HCN/molar mass of HCN
=(1.85 g)/(27.03 g/mol)
= 6.845*10^-2 mol
volume , V = 3.5 L


use:
Molarity,
M = number of mol / volume in L
= 6.845*10^-2/3.5
= 1.956*10^-2 M

HCN dissociates as:

HCN          ----->     H+   + CN-
1.956*10^-2                 0         0
1.956*10^-2-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)*1.956*10^-2) = 3.482*10^-6

since c is much greater than x, our assumption is correct
so, x = 3.482*10^-6 M



So, [H+] = x = 3.482*10^-6 M


use:
pH = -log [H+]
= -log (3.482*10^-6)
= 5.458
Answer: D