You have a solution of .5 L of 1.74g of NH3 and 4.15g of (NH4)2SO4. How many ml of 14M HCl is required for a pH of 9.00?
I don't know how to go finish this one. I thought about figuring the concentrations of each (NH3, NH4), making an ice table and solving for the additino of H+. But then i don't know how to get ml, cuz what i solved would be in conc of H+ wouldn't it? :s
(Explanation would be greatful, thanks)
Yes, you would have concn BUT M = moles/L so if you know moles and M you can calculate L and convert to mL. I think you have the right idea; you just quit too soon.
moles NH3 = 1.74/17.03 = ?
moles (NH4)2SO4 = 4.19/132.14 = ? and that x 2 = about 0.0628
...........NH3 + H^+ ==> NH4^+
initial...0.102...0.....0.0628
add...............x
change.....-x....-x.......+x
equil.....0.102-x..0.....0.0628+x
pH = pKa + log[(base)/(acid)]
9.00 = 9.25 + log [(0.102-x)/(0.0628+x)]
Solve for x. If I didn't goof that should be close to 0.0427 mols H^+.
M = moles/L
M = 14, moles = 0.0427, solve for L and convert to mL and I get approximately 3 mL but you need to be a little more accurate than that. I always like to check it to see if I get the right pH.
NH3 = 0.102-0.0427 = 0.0593
NH4^+ = 0.0628+0.0427 = 0.1055
pH = 9.25 + log (0.0593/0.1055) = 9.00
Voila.
To solve this problem, we need to approach it step by step.
Step 1: Calculate the number of moles of NH3 and (NH4)2SO4.
- Given: 1.74g of NH3 and 4.15g of (NH4)2SO4
- To find the number of moles, we use the formula: moles = mass / molar mass.
- The molar mass of NH3 is approximately 17.03 g/mol, so the number of moles of NH3 is 1.74g / 17.03 g/mol = 0.102 moles.
- The molar mass of (NH4)2SO4 is approximately 132.14 g/mol, so the number of moles of (NH4)2SO4 is 4.15g / 132.14 g/mol = 0.031 moles.
Step 2: Determine the limiting reagent.
- The reaction between NH3 and HCl produces NH4+ ions and Cl- ions, which contribute to the pH of the solution.
- We need to determine which reagent, NH3 or (NH4)2SO4, will be used up first.
- To do this, we compare the moles of NH3 and (NH4)2SO4.
- Since the ratio of NH3 to (NH4)2SO4 is 1:1 in the reaction, the limiting reagent will be the one that has fewer moles. In this case, it is (NH4)2SO4.
Step 3: Calculate the concentration of NH4+ ions.
- The number of moles of NH4+ ions is equal to the number of moles of (NH4)2SO4 used since there is a 1:1 ratio between them.
- The volume of the solution is given as 0.5 L.
- Therefore, the concentration of NH4+ ions is: moles / volume = 0.031 moles / 0.5 L = 0.062 M.
Step 4: Calculate the volume of 14M HCl required.
- Since we know that the concentration of H+ ions from the HCl solution will contribute to the pH, we need to determine how much of the HCl solution is required to achieve a pH of 9.00.
- We can use the pH formula: pH = -log[H+].
- Rearranging the formula, [H+] = 10^(-pH).
- For a pH of 9.00, [H+] = 10^(-9).
- The concentration of H+ ions from the HCl solution is equal to the concentration of HCl, which is given as 14M.
- Therefore, we can set up the equation: [H+] = 14 / V, where V is the volume of 14M HCl required.
- Rearranging the equation, V = 14 / [H+].
Substituting the known values, V = 14 / (10^(-9)). Solving this equation will give you V in liters.
To convert V to milliliters, multiply the result by 1000 since there are 1000 milliliters in a liter.
Note: The answer obtained will give you the volume of HCl solution required to achieve the desired pH. If you only have a solution with a concentration of 14M HCl, you need to dilute it accordingly to obtain the desired volume.