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.
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.
To calculate the molarity of the sodium hydroxide (NaOH) solution, we need to follow these steps:
Step 1: Determine the moles of hydrochloric acid (HCl) used in the titration.
The volume of HCl used is 22.0 cm³. Since the molarity of HCl is 0.3 mol/dm³, we can calculate the moles of HCl used using the equation:
moles of HCl = molarity × volume
moles of HCl = 0.3 mol/dm³ × 22.0 cm³
Step 2: Determine the moles of NaOH that react with the HCl.
According to the balanced chemical equation, the mole ratio between NaOH and HCl is 1:1. Therefore, the number of moles of NaOH that react is the same as the moles of HCl used.
Step 3: Convert the moles of NaOH to volume.
Since the volume of NaOH used is given as 37.0 cm³, we can calculate the molarity of NaOH using the equation:
molarity of NaOH = moles of NaOH / volume in dm³
molarity of NaOH = moles of NaOH / (37.0 cm³ ÷ 1000 cm³/dm³)
Now let's do the calculations:
Step 1:
moles of HCl = 0.3 mol/dm³ × 22.0 cm³ = 0.0066 mol
Step 2:
moles of NaOH = moles of HCl = 0.0066 mol
Step 3:
molarity of NaOH = 0.0066 mol / (37.0 cm³ ÷ 1000 cm³/dm³)
molarity of NaOH = 0.0066 mol / 0.037 dm³
molarity of NaOH ≈ 0.178 mol/dm³
Therefore, the molarity of the sodium hydroxide (NaOH) solution is approximately 0.178 mol/dm³.
To determine the molarity of the sodium hydroxide (NaOH) solution, we need to use the balanced chemical equation and stoichiometry.
From the balanced equation: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l), we can see that the ratio between NaOH and HCl is 1:1. This means that one mole of NaOH reacts with one mole of HCl.
Given:
Volume of HCl used = 22.0 cm³
Molarity of HCl solution = 0.3 mol dm⁻³
First, we need to calculate the number of moles of HCl used in the titration:
Number of moles of HCl = Molarity x Volume (in dm³)
= 0.3 mol dm⁻³ × (22.0 cm³ ÷ 1000 cm³/dm³)
= 0.0066 moles
Since the ratio between NaOH and HCl is 1:1, this means that 0.0066 moles of NaOH were also used in the titration.
Now, we can calculate the molarity of NaOH using the volume of NaOH solution:
Molarity of NaOH = Number of moles ÷ Volume (in dm³)
= 0.0066 moles ÷ (37.0 cm³ ÷ 1000 cm³/dm³)
= 0.178 mol dm⁻³
Therefore, the molarity of the sodium hydroxide (NaOH) solution is approximately 0.178 mol dm⁻³.