A chemist in a water treatment plant requires five solutions of NaOH – a stock solution and four standard solutions.
40.00 g of NaOH are dissolved in water in a 250 mL volumetric flask to prepare the stock solution.
What is the concentration in moles/litre of the stock solution? Fill in the Blank 01 M
The chemist then takes four 250 mL volumetric flasks to prepare the standard solutions. 1 mL, 2 mL, 5 mL and 10 mL of stock are pipetted into these four flasks, and the volume made up with water.
The concentrations of these four flasks are..?
(Atomic Masses; Na = 23.0, O = 16.0, H = 1.00)
40.00 g of NaOH are dissolved in water in a 250 mL volumetric flask to prepare the stock solution.
What is the concentration in moles/litre of the stock solution? Fill in the Blank 01 M
The chemist then takes four 250 mL volumetric flasks to prepare the standard solutions. 1 mL, 2 mL, 5 mL and 10 mL of stock are pipetted into these four flasks, and the volume made up with water.
The concentrations of these four flasks are..?
(Atomic Masses; Na = 23.0, O = 16.0, H = 1.00)
-
1) The atomic mass of NaOH is 23.0 + 16.0 + 1.00 = 40.0g/mol
The moles of dissolved NaOH is 40.00g of NaOH * 1mol/40g = 1 mol NaOH
1 mol of NaOH is dissolved in 250 mL water, which gives a concentration of 1mol/0.25L = 4mol/L = 4 M
2) 1mL of 4M NaOH = 0.001 L * 4mol/L = 0.004 mol NaOH is present. If the volume is diluted from 1 mL to 250 mL, the resulting concentration would be 0.004mol/0.250L = 0.016 M NaOH
The same procedure for the next three.
2mL of 4M NaOH = 0.002 L * 4mol/L = 0.008 mol NaOH is present. If the volume is diluted from 1 mL to 250 mL, the resulting concentration would be 0.008mol/0.250L = 0.032 M NaOH
5mL of 4M NaOH = 0.005 L * 4mol/L = 0.020 mol NaOH is present. If the volume is diluted from 1 mL to 250 mL, the resulting concentration would be 0.020 mol/0.250L = 0.080 M NaOH
10mL of 4M NaOH = 0.010 L * 4mol/L = 0.040 mol NaOH is present. If the volume is diluted from 1 mL to 250 mL, the resulting concentration would be 0.040mol/0.250L = 0.160 M NaOH
The moles of dissolved NaOH is 40.00g of NaOH * 1mol/40g = 1 mol NaOH
1 mol of NaOH is dissolved in 250 mL water, which gives a concentration of 1mol/0.25L = 4mol/L = 4 M
2) 1mL of 4M NaOH = 0.001 L * 4mol/L = 0.004 mol NaOH is present. If the volume is diluted from 1 mL to 250 mL, the resulting concentration would be 0.004mol/0.250L = 0.016 M NaOH
The same procedure for the next three.
2mL of 4M NaOH = 0.002 L * 4mol/L = 0.008 mol NaOH is present. If the volume is diluted from 1 mL to 250 mL, the resulting concentration would be 0.008mol/0.250L = 0.032 M NaOH
5mL of 4M NaOH = 0.005 L * 4mol/L = 0.020 mol NaOH is present. If the volume is diluted from 1 mL to 250 mL, the resulting concentration would be 0.020 mol/0.250L = 0.080 M NaOH
10mL of 4M NaOH = 0.010 L * 4mol/L = 0.040 mol NaOH is present. If the volume is diluted from 1 mL to 250 mL, the resulting concentration would be 0.040mol/0.250L = 0.160 M NaOH
-
Standard # 1 : 1 ml of stock in 250 ml water =4 ml of 1 M solution in a liter. So this is 0.004 M
Standard # 2: 2 ml of stock in 250 ml water = 8 ml of 1M solution in a liter. So this is 0.008 M
Standard # 3: 5 ml of stock in 250 ml water = 20 ml of 1M solution in a liter. So this is 0.020 M
Standard # 4: 10 ml of stock in 250 ml water= 40 ml of i M solution in a liter. So this is 0.040 M NaOH solution.
Standard # 2: 2 ml of stock in 250 ml water = 8 ml of 1M solution in a liter. So this is 0.008 M
Standard # 3: 5 ml of stock in 250 ml water = 20 ml of 1M solution in a liter. So this is 0.020 M
Standard # 4: 10 ml of stock in 250 ml water= 40 ml of i M solution in a liter. So this is 0.040 M NaOH solution.