I have gotten those numbers right so far, but I\'m not sure what they are asking
ID: 886959 • Letter: I
Question
I have gotten those numbers right so far, but I'm not sure what they are asking for in the ICE Table and the Ka= part. I thought CN would be 0 in the ICE table but apparently it's not and I don't understand why.
Compute the pH of the solution after the reaction of 53mL of 0.67M HCN with 3.0 × 101mL of 0.40 KOH HCN(aq)+KOH(aq) HOH() + KCN(aq Compute the moles of reactants before reaction Number Number Moles of acid Moles of base 03551 moles 012 moles Moles of reactant and pH active product after reaction Number Number Excess reactant 02351 moles 0.012 moles Now convert moles to concentration units Number Number Reactant] .28325 Molar 14457 Molar Set up the ICE TABLE Scroll down... HCN Number 28325 CN Number Number Initial conc +x tX Equilibrium these are text fields. Report the numbers as 0.something like 0.11 not just.11! Use the equilibrium concentrations from the ICE table and make the appropriate approximation to fill in the table below. Then solve for x and pH. Ka of HCN is 3.5x104 Number Scroll down. Number Now Compute pH Number Number pH =Explanation / Answer
ICE table :
HCN <---------------------> H+ + CN-
0.28325 0 0 ------------------------> initial
-x + x + x ---------------------> changed
0.28325 -x x x ----------------------> equilibrium
Ka = [H+][CN-]/[HCN]
Ka = x^2 / 0.28325 -x
3.5 x 10^-4 = x^2 / 0.28325 -x
x^2 + 3.5 x 10^-4 x - 9.914 x 10^-5 = 0
x = 9.78 x 10^-3
x = [H+] = 9.78 x 10^-3 M
[H+] = 9.78 x 10^-3 M
pH = -log [H+]
pH = -log ( 9.78 x 10^-3 )
pH = 2.01
note : i did not touch your soved part that is above part .