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
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?
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