This is a question I got for homework:
12.5 cm^3 of 0.5 moldm^-3 hydrochloric acid neutralises 50.0 cm^3 of a given solution of sodium hydroxide. What is the concentration of the sodium hydroxide solution in moldm^-3?
This is not just me using yahoo answers to do my homework, I was off ill and didn't get this properly explained. If I understand this I can do the rest of the questions :)
12.5 cm^3 of 0.5 moldm^-3 hydrochloric acid neutralises 50.0 cm^3 of a given solution of sodium hydroxide. What is the concentration of the sodium hydroxide solution in moldm^-3?
This is not just me using yahoo answers to do my homework, I was off ill and didn't get this properly explained. If I understand this I can do the rest of the questions :)
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1 HCL+1 NaOH -> 1 NaCL+ 1 H2O
(12.5 cm^3).(0.5 moldm^-3 )=6.25x10^-3 mol of HCL
It's a 1:1 equation
So
Mol of HCL used=Mol of NaOH used
Concentration= Mol/Volume
Concentration of NaOH= (6.25 x 10^-3mol )/ (50.0cm^3)= 0.125 Moldm^-3
You need to convert all cm into dm
(12.5 cm^3).(0.5 moldm^-3 )=6.25x10^-3 mol of HCL
It's a 1:1 equation
So
Mol of HCL used=Mol of NaOH used
Concentration= Mol/Volume
Concentration of NaOH= (6.25 x 10^-3mol )/ (50.0cm^3)= 0.125 Moldm^-3
You need to convert all cm into dm
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NaOH + HCl = NaCl + H2O
No. of moles of NaOH = No. of moles of HCl
MbVb = MaVa
Mb = (MaVa)/Vb = (0.50mol/L x 12.50mL)/50mL = 0.125mol/L = concentration of the base (NaOH).
No. of moles of NaOH = No. of moles of HCl
MbVb = MaVa
Mb = (MaVa)/Vb = (0.50mol/L x 12.50mL)/50mL = 0.125mol/L = concentration of the base (NaOH).