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 = 11.50?
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first i did was
ph = 11.50
PH= 14-11.50
= 2.5
10^-2.5
= 0.0032 M
Volume - 0.65 L
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
sooooooooooo
v1 = c2v2/c1
= 0.65 * 0.0032 / 3.88
= 52.35
52350 ml
pzl chk again
thnks
2 answers
This problem is different than the first one; however, I didn't do that one right. Give me a few seconds and we can talk about this one.
I made a mistake when I worked the first problem with a final pH = 11.55.
But this problem is done the way (and the 11.55 problem is worked the same way).
You're ok to pOH = 2.5 and (OH^-) = 0.00316 and since you have three significant figures I would not throw one away and round to 0.0032. I would leave it 0.00316 = OH^- which is what you want in the 650 mL final volume. But you haven't done the M1V1 = M2V2 correctly. You are supposed to put in (NH3) and not concn (OH^-)
The mistake I made when I did the first one is this. The NH3 is not a strong base; it is a weak base and doesn't ionize completely. Therefore, we must calculate the NH3 concn in the 650 mL final volume.
NH3 + HOH ==> NH4^ + OH^-
Kb = 1.8 x 10^-5 = (NH4^+)(OH^-)/(NH3) and solve for (NH3).
(NH3) = (NH4^+)(OH^-)/1.8 x 10^-5
and I have (NH3) = (0.00316)(0.00316)/1.8 x 10^-5 and I get (NH3) = 0.555 M.
Now we can do M1V1 = M2V2
3.88*V1 = 0.555*650
V1 = about 92 something.
Note that I did not try to make a correction for the ionization of NH3 as compared to the 0.555 M at the end. If you want to do that to see if it makes a difference, the equation would look like this.
(NH3)-0.00316 = (0.00316)(0.00316)/1.8 x 10^-5.
Solving I get (NH3) = 0.558 instead of 0.555 so it makes a slight difference. Since it doesn't involve a quadratic, that probably should be done.
It makes less than a mL difference, that is the final number is about 93 mL of the 6.8% soln to be added to enough water to make 650 mL final volume. Check my work.
But this problem is done the way (and the 11.55 problem is worked the same way).
You're ok to pOH = 2.5 and (OH^-) = 0.00316 and since you have three significant figures I would not throw one away and round to 0.0032. I would leave it 0.00316 = OH^- which is what you want in the 650 mL final volume. But you haven't done the M1V1 = M2V2 correctly. You are supposed to put in (NH3) and not concn (OH^-)
The mistake I made when I did the first one is this. The NH3 is not a strong base; it is a weak base and doesn't ionize completely. Therefore, we must calculate the NH3 concn in the 650 mL final volume.
NH3 + HOH ==> NH4^ + OH^-
Kb = 1.8 x 10^-5 = (NH4^+)(OH^-)/(NH3) and solve for (NH3).
(NH3) = (NH4^+)(OH^-)/1.8 x 10^-5
and I have (NH3) = (0.00316)(0.00316)/1.8 x 10^-5 and I get (NH3) = 0.555 M.
Now we can do M1V1 = M2V2
3.88*V1 = 0.555*650
V1 = about 92 something.
Note that I did not try to make a correction for the ionization of NH3 as compared to the 0.555 M at the end. If you want to do that to see if it makes a difference, the equation would look like this.
(NH3)-0.00316 = (0.00316)(0.00316)/1.8 x 10^-5.
Solving I get (NH3) = 0.558 instead of 0.555 so it makes a slight difference. Since it doesn't involve a quadratic, that probably should be done.
It makes less than a mL difference, that is the final number is about 93 mL of the 6.8% soln to be added to enough water to make 650 mL final volume. Check my work.
3 answers