Omg, how do i do this?

[ManBearPig]

A chemist uses a standard solution of 0.052 M sodium hydroxide (NaOH) to titrate 16.64 mL of hydrochloric acid (HCl), she finds that it requires 25.57 mL of the base to reach the end-point of the titration. What is the molarity of the acid solution ?

I thought it was m1*v1=m2*v2

but it apparantly doesnt work....any ideas?

 

[KLin]

42 gallons of HCl.

/thread

 

[HaxorNubcake]

is the answer 0.0799M?

er nvm

 

[soydios]

I took Chem last year, here's what I remember (including significant figures at each step):

(0.02557 mL OH- solution) x (0.052 molar OH- solution) = 0.0013 moles OH-
(0.0013 moles OH-) = (0.0013 moles H+)
(0.0013 moles H+) / (0.01664 mL H+ solution) = 0.078 molar H+ solution

 

[HaxorNubcake]

Originally posted by: soydios
I took Chem last year, here's what I remember (including significant figures at each step):

(0.02557 mL OH- solution) x (0.052 molar OH- solution) = 0.0013 moles OH-
(0.0013 moles OH-) = (0.0013 moles H+)
(0.0013 moles H+) / (0.01664 mL H+ solution) = 0.078 molar H+ solution

yeah same as my answer, i just kept all the numbers.

thats the same answer as m1v1=m2v2 which would make sense, but there might be some diff way to do it. I dunno chem was 2 years ago for me

 

[ManBearPig]

Hm...that doesnt seem to work either. Thanks a lot though.

 

[Duddy]

4 tons of TNT.

 

[DAWeinG]

3.15x10^-2 ? *shrug*

 

[ManBearPig]

it was .080. damn sig figs.

 

[gsethi]

50mL of 0.2mol L-1 NaOH neutralised 20mL of sulfuric acid. Determine the concentration of the acid

1. Write the balanced chemical equation for the reaction
NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2H2O(l)
2. Extract the relevant information from the question:
NaOH V = 50mL, M = 0.2M H2SO4 V = 20mL, M = ?
3. Check the data for consistency
NaOH V = 50 x 10-3L, M = 0.2M H2SO4 V = 20 x 10-3L, M = ?
4. Calculate moles NaOH
n(NaOH) = M x V = 0.2 x 50 x 10-3 = 0.01 mol
5. From the balanced chemical equation find the mole ratio
NaOH:H2SO4
2:1
6. Find moles H2SO4
NaOH: H2SO4 is 2:1
So n(H2SO4) = ½ x n(NaOH) = ½ x 0.01 = 5 x 10-3 moles H2SO4 at the equivalence point
7. Calculate concentration of H2SO4: M = n ÷ V
n = 5 x 10-3 mol, V = 20 x 10-3L
M(H2SO4) = 5 x 10-3 ÷ 20 x 10-3 = 0.25M or 0.25 mol L-1

Taken from: http://www.ausetute.com.au/titrcalc.html
Example 2

 

[HaxorNubcake]

Originally posted by: Heen05
it was .080. damn sig figs.

yay so I was correct...

 

[Duddy]

Holy Christ some of you guys are WAY too smart!

 

[JustAnAverageGuy]

Originally posted by: Duddy
Holy Christ some of you guys are WAY too smart!

Don't worry. I hated chem, even in high school

- JaAG

 

[Duddy]

Holy Christ some of you guys are WAY too smart!

 

[bonkers325]

looks like someone forgot to do their chem hw over the weekend!