Calculate the pH of the solution that is prepared by mixing 26.06g of ammonium iodide (MM=144.942) and 5.08g of ammonia (MM=17.034g) in enough 0.066 M HCI to produce 1.00L of solution. Kb(NH3)=1.76x10^-5

I converted 24.06g and 5.08g to moles and i now I have to set up an ice box but i'm not sure how to write the reaction. I had NH4--> H+ + NH3 but I know that's not right. After the equation I would have to set up an ice box and solve for Ka then take the -log or that to find the pH?

This is a buffer problem with the slight complication of added HCl.

The NH4I + NH3 is a buffer solution and the pH is determined by the Henderson-Hasselbalch equation. The complication is that HCl has been added to it so the thing to do is to calculate mols NH3 and subtract the mols HCl so that you are left with mols NH4I and mols NH3 (after the neutralization. But the neutralization forms NH4Cl and mols NH4Cl added to the mols NH4I will be the acid part of the H-H equation. mols NH3 will be the base part. Then pH = pKa + log(base)/(acid) and that should get it.
NH3 + HCl ==> NH4Cl

To determine the pH of the solution, you need to set up an ice box and consider the reaction between ammonium ion (NH4+) and water (H2O) according to the following equation:

NH4+ + H2O ⇌ NH3 + H3O+

To find the concentration of the ammonium ion (NH4+), begin by calculating the moles of ammonium iodide.

Moles of ammonium iodide = mass / molar mass
Moles of ammonium iodide = 26.06g / 144.942 g/mol

Next, calculate the concentration of the ammonium ion using the volume of the final solution:

Concentration of NH4+ = moles of NH4+ / volume of solution (L)

Since the final volume is given as 1.00 L, you can now determine the concentration.

Concentration of NH4+ = (moles of NH4+) / 1.00 L

Repeat the same steps for ammonia (NH3) using its given mass and molar mass:

Moles of ammonia = mass / molar mass
Moles of ammonia = 5.08g / 17.034g/mol

Concentration of NH3 = (moles of NH3) / 1.00 L

Once you have obtained the concentrations of NH4+ and NH3, you can set up the equilibrium expression:

Kb = [NH3][H3O+] / [NH4+]

Since Kb is given as 1.76x10^-5, you can rearrange the equation to solve for [H3O+]:

[H3O+] = (Kb x [NH4+]) / [NH3]

Evaluate the concentrations and solve for [H3O+]. Finally, calculate the pH using the equation:

pH = -log[H3O+]

Remember to round your calculated pH value to the appropriate number of decimal places.