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
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?
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i am gettig 5.3 x 10^-10
is that rite or worng
pzl conform ..
3 answers
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
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
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.
3 answers