determining unknown concentrations (1 Viewer)

[n@ttyb]

a 5.00 ml sample NAOH was neutralised by 20.52 ml of a 0.952 m L-1 HCL.
find conc. of NAOH.

i worked it out as following:

n(HCL) = c*v
= 0.952 * 0.02052
=0.0195
n(NAOH)= n(HCL)
C(NAOH) = n/v
= 0.0195/0.00500
=3.9 mol L-1

Is that 3.9 possible ?
All the others i have done are in the decimals.. have i done something wrong in the working out ?

 

[Riviet]

Yep, seeing as it took 4 times as much HCl of (almost 1 molar) to neutralise the NaCl, it must have been strong.

 

[tristambrown]

from 5ml of Unknown conc NaOH neutralised by 20.52 ml of 0.952 molar HCL i am getting 3.907008
~3.9

there is nothing wrong with your working