The equation for the reaction between sodium hydroxide and hydrochloric acid is
NaOH + HCl -> NaCl + H20
Relative formula masses:
NaOH- 40, HCl - 35.5
Some hydrochloric acid was added to 25.0cm cubed of NaOH solution during titration. The sodium hydroxide was just neutralised when 23.0cm cubed of acid was added. The concentration of the acid was 0.365g/dm cubed.
What mass of HCl neutralised the sodium hydroxide?
Using mass= V x C I did 23.0cm cubed divided by 1000 to make it dm cubed. then 0.023 x 0.365 g/dm cubed. This gave me 8.395 x10 to the power of -3.
a) What mass of NaOH reacts with the mass of HCl in your answer to the previous question?
My teacher said this is a 'reacting masses calc' but after doing this I got an answer of
0.009459. Is this the correct answer or has it gone wrong somewhere?
Thanks
Sorry for the long explanation!
NaOH + HCl -> NaCl + H20
Relative formula masses:
NaOH- 40, HCl - 35.5
Some hydrochloric acid was added to 25.0cm cubed of NaOH solution during titration. The sodium hydroxide was just neutralised when 23.0cm cubed of acid was added. The concentration of the acid was 0.365g/dm cubed.
What mass of HCl neutralised the sodium hydroxide?
Using mass= V x C I did 23.0cm cubed divided by 1000 to make it dm cubed. then 0.023 x 0.365 g/dm cubed. This gave me 8.395 x10 to the power of -3.
a) What mass of NaOH reacts with the mass of HCl in your answer to the previous question?
My teacher said this is a 'reacting masses calc' but after doing this I got an answer of
0.009459. Is this the correct answer or has it gone wrong somewhere?
Thanks
Sorry for the long explanation!
-
strength of HCl soln. = 0.365/36.5=0.01
mmoles used in titration= 0.01 x 23 = 0.23 and mass = 0.23 x 36.5 / 1000 = 0.008395 g
ans. is correct
mmole of NaOH = mmole HCl = 0.23
mass = 0.23 x 40 /1000 = 0.0092 g
mmoles used in titration= 0.01 x 23 = 0.23 and mass = 0.23 x 36.5 / 1000 = 0.008395 g
ans. is correct
mmole of NaOH = mmole HCl = 0.23
mass = 0.23 x 40 /1000 = 0.0092 g