Question

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 am gettig 5.3 x 10^-10

is that rite or worng
pzl conform ..

Answer 1

No. I don't get that answer.
What did you get for the molarity of the 6.8% NH3 solution? I have 3.88 M.
For a solution to be pH = 11.55, what do you have for the (OH^-)? I have 0.00355 M

Answer 2

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

Answer 3

See my response to your second post above and note that I didn't work this problem correctly. I have corrected it there, or at least I worked the other problem correctly and this one follows the same procedure.