I think the first thing you need to do is to write ionization equations and calculate the molarity of each of the components. For the solid, you divide the grams by molar mass NH4I to determine mols and divide that by vol (in Liters) to convert that to molarity. If I didn't goof NH4I is 1 M.
Then remembering NaOH is a strong base and ionizes 100% and NH3 is a weak base with a Kb (relatively small) PLUS the NH4I will act as a common ion (the NH4 from NH4I is common to the NH4^+ produced by NH3 + HOH ==> NH4^+ + OH^-
The majority of the NH4^+ will come from the NH4I with a small amount from NH3 reaction with water. You may want to calculate that (or at least estimate it) to see if the amount is significant. The majority of the OH^- will come from NaOH (also NaOH is a common ion to the NH3 reaction, also, which reduces the contribution of the NH3 reaction both to the NH4^+ as well as to the OH^-). I didn't go through the calculation but I suspect the NH4^+ and the OH^- from the NH3 + HOH reaction will be VERY small, and possibly negligible.
To a 100 mL volumetric flask were added 5.00 mL of 6.0 M NH3 , 14.5g of solid ammonium iodide, and 40.0 mL of 0.50 M NaOH. The flask was filled to the calibration mark with water.
(a) What were the molar concentrations of hydrogen ion, hydroxide ion, and ammonium ion in the resultant solution? K = 1.8 x 10-5 for NH3
what do you do with the solid in terms of concentration? How do you determine the dominant equation?
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