Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: govibe on January 06, 2008, 06:20:31 PM
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"Calculate the pH after 0.020 mol NaOH is added to 1.00 L of each of the following solutions.
(a) 0.480 HONH2 (Kb = 1.1x10-8)
(b) 0.480 M HONH3Cl
(c) a mixture containing 0.480 M HONH2 and 0.480 M HONH3Cl"
for (a), the reaction, the dissociation is HONH2+ H2O --->>> H4ON+ + OH- ??
The ICE box is like
HONH2+ H2O --->>> H4ON+ + OH-
0.480M 0 0.02
-x +x +x
so the Kb=1.1E-8 = (0.02+x)x/(0.480-x)
so x is 2.64E-7, so OH- concentration is 0.02+2.64E-7=0.02. The negative log of that is 1.7, and 14-1.7 is 12.30, is that correct? Do I have to do a stoichiometry equation in eddition to an ICE box?
for parts b and c, I'm not sure the dissociation of HONH3Cl is...
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So your main problem is whether hydroxylamine is basic or acidic... It is basic, it can get protonated like ammonia.
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what does the salt HONH3Cl dissociate into?