Question help, Thanks!

[zibb3r]

Hey guys, I have got a couple more question, if you guys could help, that would be great!

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:

[luken93]

1) Work backwards

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

----------------

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

[zibb3r]

Thanks! :smitten:

[luken93]

Thanks! :smitten:

Are they correct? I'm pretty sure they are but I wouldn't want to mislead you haha

[m@tty]

^It all looks alright.

Except for the original equation:

??

It should be one water, not two yes?

[luken93]

^It all looks alright.

Except for the original equation:

??

It should be one water, not two yes?

Am I tripping or is there 2x H's and 4x O's on the LHS?

[m@tty]

You kept the nitrate ion as is though.. So you get the H and the oxide for only one water molecule.

[luken93]

You kept the nitrate ion as is though.. So you get the H and the oxide for only one water molecule.

\facepalm

[m@tty]

Good thing is it had no effect on the actual question.