A particular household ammonia solution ( d= 0.97 ) is 6.8% NH3 by mass
How many milliliters of this solution should be diluted with water to produce 600 ml of a solution pH= 11.65?
First you need to determine the concentration of NH3 in the final solution.
NH3 + HOH ==> NH4^+ + OH^-
Kb = (NH4^+)(OH^-)/(NH3)
You will need to look up Kb, if it isn't given in the problem, then plug in for OH^- and NH4^+ the OH^- you obtain from a pH of 11.65. That means a pOH of 2.35 and a (OH^-) of about 0.00447 but you need to go through it exactly since I've estimated. (Also, you must determine if the NH3 to use is NH3 - OH or just NH3. The size of the numbers will help you determine that.
Second, you need to determine the molarity of the NH3 solution you have (that's the 6.8% solution).
It has a density of 0.97 g/mL; therefore, 1000 mL will have a mass of
0.97 g/mL x 1000 mL = 970 grams.
How much of that is NH3. That will be
970 grams x 0.068 = 66 g NH3.
How many moles is that. grams/molar mass = moles = 66/17 = approximately 4 M.
Now use M(soln 1) x L(soln 1) = M(soln 2) x L(soln 2).
Post your work if you get stuck. And note that I've estimated at each step along the say so you need to redo all of this but use the exactly values.
Ok here is my solution to it but i am constantly told that it is wrong.
d=0.97g/mL
Mass %= 6.8%
1000mL= 1L
d=m/v m=d X v
m= 0.97g/mL X 1000mL
m=970g
mNH3 = 970g X 0.068=65.96g of NH3
MM of NH3 = 14g/mol + 3 X 1g/mol=17g/mol
nNH3 = 65.96g/17g/mol
n=3.88mol
[NH3]= 3.88mol/L or M
pH=11.65 thus pOH=2.35 thus [OH-]=4.466X10^-3
V=0.625L
Now using the ICE table for NH3 + H2O =(equilibrium sign) NH4^+ + OH^-
Kb=1.80 X 10^-5
I 3.88 0 0
C -x +x +x
E 3.88-x x x
Kb= [NH4^+][OH^-]/[NH3]
1.80 X 10^-5= x^2/3.88-x
neglecting the x because Kb is very small( I hope i am right)
1.80 X 10^-5 X 3.88 = x^2
using calculator and solving I got
x=8.3570X10^-3
similary X is [OH-]= 8.3570X10^-3
Using c1v1=c2v2
v1=c2v2/c1
v1=4.466X10-3 X 0.625/8.3570X10^-3
v1= 0.334L
which is 334mL and is the final answer.
But why in the NAME OF ARHENIUS, is the my answer not right.
DR. BOBB or any chem genius could tell me if I am missing anything becuase I took half hour to type the solution online. That is the how serious I am about this question.
and forgot to mention, I need a volume of 625mL of pH of 11.65....differnt from the original question posted....
To determine the amount of the particular household ammonia solution that should be diluted with water to produce a pH 11.65 solution, we need to consider the concentration of NH3 in the original solution and the desired pH of the final solution.
Here's how you can calculate it:
Step 1: Find the concentration of NH3 in the original solution.
The original household ammonia solution is 6.8% NH3 by mass. This means that for every 100g of the solution, there are 6.8g of NH3.
Step 2: Convert the mass of NH3 to moles.
We need to convert the mass of NH3 to moles by using the molar mass of NH3, which is approximately 17.03 g/mol.
moles of NH3 = (mass of NH3 / molar mass of NH3)
Step 3: Determine moles of NH4OH in the desired solution.
The desired solution has a pH of 11.65, which is basic. In basic solutions, NH3 acts as a base and forms NH4OH with water.
To calculate the exact amount of NH4OH in moles needed, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA]),
where:
pH = 11.65 (desired pH)
pKa = 9.24 (pKa of NH4OH, which is equal to pKb of NH3)
[HA] = concentration of NH3 (moles/L)
[A-] = concentration of NH4OH (moles/L)
Since we are given the desired pH and pKa values, and we need to find [A-], we rearrange the Henderson-Hasselbalch equation to solve for [A-]:
[A-] = 10^(pH - pKa) * [HA]
Substituting the values:
[A-] = 10^(11.65 - 9.24) * [HA]
Step 4: Determine the moles of NH3 in the final solution.
Since NH4OH is formed by reaction with NH3, the moles of NH3 must be equivalent to the moles of NH4OH in the final solution. Therefore, the moles of NH3 required are equal to [A-].
Step 5: Convert the moles of NH3 to mass.
mass of NH3 = (moles of NH3 * molar mass of NH3)
Step 6: Find the volume (in mL) of the household ammonia solution.
Knowing that the density (d) of the household ammonia solution is 0.97 g/mL and the mass of NH3 required, we can calculate the volume of the original solution as:
volume of household ammonia solution = (mass of NH3 required / density of the solution)
Step 7: Calculate the volume of water needed to dilute the solution.
The total volume of the final solution will be 600 mL. Therefore, the volume of water needed can be calculated as:
volume of water = (total volume - volume of household ammonia solution)
By following these steps, you should be able to determine the volume of the particular household ammonia solution that should be diluted with water to produce a 600 mL solution with a pH of 11.65.