# urgent chem help ( dr bob plz)

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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

• urgent chem help ( dr bob plz) -

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.

• urgent chem help ( dr bob plz) -

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.

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