200cm^3 of 1.0 mol dm^-3 sulphuric acid,H2SO4 is poured into a 250 cm^3 volumetric flask.Distilled water is then added to make 250 cm^3 of solution.
(a)What is the molarity of the diluted acid solution?
(b)What is the volume of 1.6 mol dm^-3 sodium hydroxide,NaOH solution needed to neutralise 25 cm^3 of the diluted acid?
C1V1=C2V2
(a)
1.0 mol/L * 0.2L = C2 * 1.0 L
Solve for C2
(b)
C2 (from a) * 25 cm³ = (1.6 mol/L)*V2
Solve for V2.
To find the molarity of the diluted acid solution, you can use the formula:
M1V1 = M2V2
Where:
M1 = Molarity of the original acid solution
V1 = Volume of the original acid solution
M2 = Molarity of the diluted acid solution
V2 = Volume of the diluted acid solution
(a) So, let's plug in the values we have:
M1 = 1.0 mol dm^-3 (given)
V1 = 200 cm^3 (given)
M2 = ?
V2 = 250 cm^3 (given)
Now we can rearrange the formula to solve for M2:
M2 = (M1 × V1) / V2
M2 = (1.0 mol dm^-3 × 200 cm^3) / 250 cm^3
To convert cm^3 to dm^3, we divide by 1000:
M2 = (1.0 mol dm^-3 × 0.2 dm^3) / 0.25 dm^3
Simplifying the equation:
M2 = 0.8 mol dm^-3
Therefore, the molarity of the diluted acid solution is 0.8 mol dm^-3.
(b) To find the volume of sodium hydroxide solution needed to neutralize 25 cm^3 of the diluted acid, we can use the formula:
M1V1 = M2V2
Where:
M1 = Molarity of sodium hydroxide solution
V1 = Volume of sodium hydroxide solution needed
M2 = Molarity of the diluted acid solution (0.8 mol dm^-3)
V2 = Volume of the diluted acid solution (25 cm^3)
Now, let's plug in the values we have and solve for V1:
M1 × V1 = M2 × V2
(1.6 mol dm^-3) × V1 = (0.8 mol dm^-3) × (25 cm^3)
To convert cm^3 to dm^3, we divide by 1000:
(1.6 mol dm^-3) × V1 = (0.8 mol dm^-3) × (0.025 dm^3)
Simplifying the equation:
1.6 × V1 = 0.02
V1 = 0.0125 dm^3
Therefore, the volume of the sodium hydroxide solution needed to neutralize 25 cm^3 of the diluted acid is 0.0125 dm^3.