Standardization of NaOH
1. Find volume of NaOH used to reach endpoint (volume final - volume initial)
2. Using weight of KHP determine moles of KHP used. Since there is a 1:1 ratio of KHP to NaOH the moles used of NaOH will be the same. KHP = 204.23 g/mol
.450 gKHP x 1mol KHP / 204.23 gKHP x 1mol NaOH / 1mol KHP = mol NaOH
3. Divide moles of NaOH by volume of NaOH and this will yield the Molarity of NaOH.
How would you prepare .250L of a .125M Solution of Sodium Hydroxide?
x Molarity = y moles / z liters ; y = x * z --->
.250L * .125M = .03125 mol
.03125 mol * 40 grams / mol = 1.25 grams of NaOH needed then diluted to .250 L
1) Make a 250mL of a 0.2M solution of HCl using 6M HCl
Molarity1 * Volume1 = Molarity2 * Volume2
6M * X = 250mL * 0.2M
x = 8.333 mL
Since you could not accurately measure 8.333 you ended up using 8.0 mL. What is the true Molarity of the HCl solution?
6M x 8.0mL = X * 250mL
X = 48 / 250 = 0.192M
2) Exactly 25 ml of an HCl solution required 39.34 ml of 0.1013 M NaOH to reach an end point with Phenolphthalein. Calculate the molarity of HCl.
39.34 * 0.1013 / 25 = .159 M
3) You go past your endpoint what should you do? Add a measured amount of the HCl (5mL) to your Erlenmeyer flask and change the above calculation (#2) to include this added amount (now you used 30 ml of HCl instead of 25 ml). Continue to titrate with the NaOH, but go slowly so you do not go past the endpoint again.
Acid-Base Titration Lab
/uploads/1/3/6/6/13668504/hcl_naoh_lab.pdf
1. Find volume of NaOH used to reach endpoint (volume final - volume initial)
2. Using weight of KHP determine moles of KHP used. Since there is a 1:1 ratio of KHP to NaOH the moles used of NaOH will be the same. KHP = 204.23 g/mol
.450 gKHP x 1mol KHP / 204.23 gKHP x 1mol NaOH / 1mol KHP = mol NaOH
3. Divide moles of NaOH by volume of NaOH and this will yield the Molarity of NaOH.
How would you prepare .250L of a .125M Solution of Sodium Hydroxide?
x Molarity = y moles / z liters ; y = x * z --->
.250L * .125M = .03125 mol
.03125 mol * 40 grams / mol = 1.25 grams of NaOH needed then diluted to .250 L
1) Make a 250mL of a 0.2M solution of HCl using 6M HCl
Molarity1 * Volume1 = Molarity2 * Volume2
6M * X = 250mL * 0.2M
x = 8.333 mL
Since you could not accurately measure 8.333 you ended up using 8.0 mL. What is the true Molarity of the HCl solution?
6M x 8.0mL = X * 250mL
X = 48 / 250 = 0.192M
2) Exactly 25 ml of an HCl solution required 39.34 ml of 0.1013 M NaOH to reach an end point with Phenolphthalein. Calculate the molarity of HCl.
39.34 * 0.1013 / 25 = .159 M
3) You go past your endpoint what should you do? Add a measured amount of the HCl (5mL) to your Erlenmeyer flask and change the above calculation (#2) to include this added amount (now you used 30 ml of HCl instead of 25 ml). Continue to titrate with the NaOH, but go slowly so you do not go past the endpoint again.
Acid-Base Titration Lab
/uploads/1/3/6/6/13668504/hcl_naoh_lab.pdf