First, calculate the number of moles of HCl used:
0.3 mol dm-3 x 0.022 dm3 = 0.0066 moles
Since NaOH and HCl react in a 1:1 ratio, the number of moles of NaOH used is also 0.0066 moles.
To calculate the molarity of the NaOH solution, divide the number of moles by the volume in liters:
Molarity = moles/volume
Volume = 37.0 cm3/1000 cm3/1 dm3 = 0.037 dm3
Molarity = 0.0066 moles/0.037 dm3 = 0.178 mol dm-3
Therefore, the molarity of the sodium hydroxide solution is 0.178 mol dm-3.
37.0cm³ of a sodium hydroxide (NaOH) solution was pipetted into a conical
flask and titrated with a standard solution of 0.3 mol dm-3 hydrochloric acid
according to the following equation:
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
A universal indicator solution was used for the titration and it was found that
22.0 cm3 of the acid (HCl) was required to neutralise the alkali (NaOH).
Calculate the molarity of the sodium hydroxide.
1 answer