A particular household ammonia solution (d= 0.97g/mL) is 6.8% NH3 by mass.
How many milliliters of this solution should be diluted with water to produce 650 mL of a solution with pH = 11.55?
I showed you how to do the other one almost step by step. This one should be done the same way. The only difference is that this one is ammonia and the other was KOH.
ok thank u
umm i tried it,,yet im getting the wrong answer..
this is wht i did:
M1 X V1= M2 X V2
V1 = M2 X V2
----------
M1
= 2.45 X 0.65
------------
0.65
= 259.365 ml
M1*V1= M2*V2 is ok except you haven't calculated the molarity of the 6.8% solution of NH3. Do that first, then calculate the molarity of the solution you want, then you can use the dilution formula.
I DON'T GET WHATS DONE..
DR.Bob22 - could u please explain this again..thank u
it should be approx 0.6 mL
i'm not too sure if it's right though, let me know
I think it's somewhere between 0.56 mL and 0.62 mL, hope that helps!
To solve this problem, we need to use the concept of dilution and calculate the amount of the household ammonia solution that needs to be mixed with water.
Let's start solving the problem step by step:
Step 1: Calculate the mass of ammonia in 650 mL of the desired solution:
Given that the solution has a pH of 11.55, we can assume that all of the ammonia is in the form of NH3, which is a weak base. The molar mass of NH3 is 17.03 g/mol.
Using the equation pH = pKa + log([A-]/[HA]), we can determine the pKa for NH4OH, which is the acid form of ammonia. The pKa for NH4OH is 9.25.
To find the concentration of NH3 (A-) in the solution, we can assume that [NH3] = [A-].
pH = pKa + log([A-]/[NH4OH])
11.55 = 9.25 + log([A-]/[NH4OH])
2.3 = log([A-]/[NH4OH])
Now we can calculate the ratio [A-]/[NH4OH] using the logarithmic properties.
[A-]/[NH4OH] = 10^(2.3)
[A-]/[NH4OH] = 199.53
Since the solution is 6.8% NH3 by mass, we can assume that the mass of NH3 is 6.8% of the total mass of the solution.
Mass of NH3 = 0.068 (mass of the solution)
Step 2: Calculate the total mass of the solution needed:
To find the total mass of the solution required, we can divide the mass of NH3 by its concentration (density) to get the volume in milliliters.
Mass of the solution = Mass of NH3 / Concentration (density)
Mass of the solution = (0.068) (mass of the solution) / (0.97 g/mL)
Now, rearrange the equation to solve for the mass of the solution:
(mass of the solution) = (0.068) (mass of the solution) / (0.97 g/mL)
(0.97 g/mL) (mass of the solution) = (0.068) (mass of the solution)
(0.97 g/mL) = 0.068
Since the units of mass cancel out on both sides, we can solve for the mass of the solution:
mass of the solution = (0.068) / (0.97 g/mL)
mass of the solution = 0.070 mL
So, the total mass of the solution needed is 0.070 mL.
Step 3: Calculate the volume of water needed to dilute the solution:
To calculate the volume of water needed, subtract the volume of the solution from the desired total volume.
Volume of water = Total volume - Volume of solution
Volume of water = 650 mL - 0.070 mL
Volume of water = 649.93 mL
So, approximately 649.93 mL of water should be diluted with the solution to produce a 650 mL solution with a pH of 11.55.