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August 27, 2014

Homework Help: urgent chem help ( dr bob plz)

Posted by javeya on Tuesday, February 17, 2009 at 10:15pm.

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

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