Determine the pH of 0.100 M aqueous HCN to two decimal places, and determine the concentration of CN- at equlibrium to two singificant figures. Can you show me all the steps to this question? Please and thank you!

...........HCN + H2O ---> H3O^+ + CN^-

initial....0.100...........0.......0
change.....-x..............+x......+x
final...0.100-x.............x.......x

Ka = (H3O^+)(CN^-)/(HCN)
Now substitute into Ka expression the values from the ICE chart.
Ka for HCN is 7.2E-4 from an old book of mine but you need to use the value found in your text/notes.

7.2E-4 = (x)(x)/(0.1-x)
Solve for x = (H3O^+) and convert to pH.
(CN^-) is the same.
NOTE: If you solve the EXACT equation I have written above you must solve a quadratic equation. Most calculators do that with no problem.
Post your work if you get stuck.

So is Ka water H2o? not sure where you got 7.20 in my chart 7.2 goes with H2PO4^-

there is nothing for (H3O+) but I got 4.69 for CN- and 9.31 for HCN
I am so lost...

If you re-read my post I told you where I obtained 7.2E-4. Since you didn't include it in your post I had to look in one of my old texts BUT I told you to look in your text and use the value you found there.

What do you mean there is nothing for H3O^+. That's what you are calculating. And HCN can't possibly be 9.31. It was 0.100 in the problem.

To determine the pH of 0.100 M aqueous HCN and the concentration of CN- at equilibrium, we will need to consider the dissociation of HCN in water. HCN is a weak acid that partially ionizes in water to form H+ and CN-. We can use the acid dissociation constant (Ka) to calculate the pH and the equilibrium expression to determine the concentration of CN-.

Step 1: Write the balanced chemical equation for the dissociation of HCN:
HCN ⇌ H+ + CN-

Step 2: Write the equilibrium expression for the dissociation of HCN:
Ka = [H+][CN-] / [HCN]

Step 3: Given that the initial concentration of HCN is 0.100 M, we can assume that at equilibrium, the concentration of H+ will be equal to the concentration of CN-, so we can represent it as x:
HCN ⇌ x + x

Step 4: Substitute the equilibrium concentrations into the equilibrium expression:
Ka = (x)(x) / (0.100 - x)

Step 5: Make the assumption that x is a small value compared to 0.100, so we can ignore it when subtracting from 0.100:
Ka = x² / 0.100

Step 6: Given that the value of Ka for HCN is 6.2 x 10^-10, we can substitute this value into the equation and solve for x:
6.2 x 10^-10 = x² / 0.100

Step 7: Rearrange the equation to solve for x:
x² = (6.2 x 10^-10) * 0.100
x² = 6.2 x 10^-11

Step 8: Take the square root of both sides to solve for x:
x = √(6.2 x 10^-11)

Step 9: Calculate the value of x using a calculator:
x ≈ 7.87 x 10^-6

Step 10: Since the concentration of H+ is equal to the concentration of CN-, the concentration of CN- at equilibrium will also be 7.87 x 10^-6 M (rounded to two significant figures).

Step 11: To calculate the pH, we can use the equation:
pH = -log[H+]

Step 12: Substitute the concentration of H+ into the equation:
pH = -log(7.87 x 10^-6)

Step 13: Calculate the pH using a calculator:
pH ≈ 5.10

Therefore, the pH of 0.100 M aqueous HCN is approximately 5.10, and the concentration of CN- at equilibrium is approximately 7.87 x 10^-6 M.