Aqueous silver nitrate reacts with aqeous potassium iodide in a double-replacement reaction to produce a precipitate of silver iodide. If 34.6 ml of 0.563 M silver nitrate are used with 148.4 ml of potassium iodide:
a) What molarity of potassium iodide will be needed to consume all of the silver nitrate?
b) What will be the molarity of the aqueous product?
Please show your work so I can do similar problems later on
Thank you so much, I really appreciate your help!
moles AgNO3 = M x L = 0.563M x 0.0346L = 0.020 moles approximately but you need to redo all of these since I estimate all of the numbers.
KI(aq) + AgNO3(aq) ==>AgI(s) + KNO3(aq)
Now convert moles AgNO3 to moles KI using the coefficients in the balanced equation.
moles AgNO3 x (1 mole KI/1 mole AgNO3) = 0.02 x (1/1) = 0.02 mol KI.
MKI = moles KI/L KI
You know moles KI and L KI, solve for M KI.
b) The aqueous product is KNO3. Use EITHER KI or AgNO3 (you don't need both) to convert to moles KNO3. The same procedure is used with the coefficients.
moles AgNO3 x (1 mole KNO3/1 mole AgNO3) = 0.02 x (1 mole KNO3/1 mole AgNO3) = 0.02 x (1/1/) = 0.02 moles KNO3.
Then M KNO3 = moles KNO3/L soln.
YOu know moles KNO3 and you know the total volume of the solution.
Sorry, but how did you find the total volume of the solution?
34.6 mL AgNO3 + 148.4 mL KI = ? mL total.
Don't get intimidated because these are mL or some unit you don't usually see. You add a gallon to a gallon and you have 2 gallons. Or a quart to a pint and you have 3 pints.
Oh! Okay thank you, that's very good advice. I understand now
To answer these questions, we need to use the concept of Stoichiometry and the balanced chemical equation of the reaction.
The balanced chemical equation for the reaction between silver nitrate (AgNO3) and potassium iodide (KI) is:
AgNO3 + KI → AgI + KNO3
Now let's solve the question step by step:
a) What molarity of potassium iodide will be needed to consume all of the silver nitrate?
Step 1: Calculate the amount of silver nitrate used:
We are given that 34.6 ml of 0.563 M silver nitrate is used. To find the amount in moles, we use the formula:
Amount (in moles) = Concentration (in M) x Volume (in L)
So, the amount of silver nitrate used is:
0.563 M x 0.0346 L = 0.01942 moles
Step 2: Use the balanced equation to find the moles of potassium iodide (KI) needed:
From the balanced equation, we know that the stoichiometric ratio between AgNO3 and KI is 1:1. This means that for every 1 mole of AgNO3, we need 1 mole of KI.
Therefore, the moles of KI needed would be 0.01942 moles
Step 3: Calculate the volume of potassium iodide in liters:
We are given the volume of KI as 148.4 ml. To convert this to liters, we divide by 1000:
Volume (in L) = 148.4 ml / 1000 = 0.1484 L
Step 4: Calculate the molarity of potassium iodide (KI):
We can use the formula:
Molarity (M) = Amount (in moles) / Volume (in L)
Molarity of KI = 0.01942 moles / 0.1484 L ≈ 0.1309 M
Therefore, the molarity of potassium iodide needed to consume all of the silver nitrate is approximately 0.1309 M.
b) What will be the molarity of the aqueous product?
From the balanced equation, we can see that silver iodide (AgI) is the precipitate formed. This means that it will be in the solid form and will not contribute to the molarity of the solution.
Therefore, the molarity of the aqueous product will be zero.
To summarize:
a) The molarity of potassium iodide needed to consume all of the silver nitrate is approximately 0.1309 M.
b) The molarity of the aqueous product is zero since silver iodide precipitates.