Post by: zibb3r on February 07, 2011, 08:51:40 pm
1) A 10.0 ml sample of HNO3 was diluted to a volume of 100.0 ml. 25 ml of the dilute solution was needed to neutralise 50.0ml of a 0.60 M KOH solution. What was the conc. of the original nitric acid???
2)What volume of .124M HCl is needed to neutralise 2.00g of Ca(OH)2??
Thanks! :smitten:
Post by: luken93 on February 07, 2011, 08:57:23 pm
HNO3 + KOH --> KNO3 + H2O
1:1 Ratio for HNO3 and KOH
n(KOH) = cV = .6 x .05 = 0.03 mol
n(HNO3) = n(KOH) = 0.03mol
So in 25 ml we have 0.03 mol
c(HNO3) = n/V = 0.03 / 0.025 = 1.2 mol/L
Therefore, if 25ml had 0.03 mol, then 100mL has 0.12 mol
and finally, use c1V1 = c2V2
c1 = ?, V1 = 0.01L, c2 = 1.2M, V2 = 0.1L
0.01x = 0.12
x = 12M concentration of original solution
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2) Firstly make the equation:
2HCl + Ca(OH)2 = CaCl2 + 2H2O
Then, find n(Ca(OH)2)
n(Ca(OH)2) = m/M = 2/74.1 = 0.027 mol
n(HCl) = 2 x n(Ca(OH)2) = 0.054 mol
Finally, use concentration equation c = n/V
V = n/c = 0.054/.124 = 0.435L
= 435 mL
Post by: zibb3r on February 07, 2011, 09:56:49 pm
Post by: luken93 on February 07, 2011, 10:08:39 pm
Thanks! :smitten:Are they correct? I'm pretty sure they are but I wouldn't want to mislead you haha
Post by: m@tty on February 07, 2011, 10:13:48 pm
Except for the original equation:
It should be one water, not two yes?
Post by: luken93 on February 07, 2011, 10:18:43 pm
^It all looks alright.Am I tripping or is there 2x H's and 4x O's on the LHS?
Except for the original equation:??
It should be one water, not two yes?
Post by: m@tty on February 07, 2011, 10:22:02 pm
Post by: luken93 on February 07, 2011, 10:23:22 pm
You kept the nitrate ion as is though.. So you get the H and the oxide for only one water molecule.\facepalm
Post by: m@tty on February 07, 2011, 10:26:53 pm