Work these titration calculations:
1. A titration of 15.0 cm3 of household ammonia, NH3, required 38.57 cm3 of 0.78M HCl. Calculate the molarity of the ammonia.
2. What volume of 0.5M HNO3 is required to neutralize 25 cm3 of a 0.2M NaOH solution?
3. Calculate the volume of 0.6M HNO3 necessary to neutralize 28.55 cm3 of 0.45M KOH.
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To solve these titration calculations, we need to use the concept of stoichiometry and the equation for the reaction in question. In these examples, we will be looking at acid-base titration reactions.
1. In the first question, the equation for the reaction between ammonia and hydrochloric acid is: NH3 + HCl -> NH4Cl.
First, calculate the number of moles of HCl used:
Volume of HCl (in L) = 38.57 cm3 * (1 L / 1000 cm3) = 0.03857 L
Number of moles of HCl = Molarity * Volume in L = 0.78 M * 0.03857 L = 0.03009 mol
Since the balanced equation is 1:1 between HCl and NH3, the number of moles of HCl used is also the number of moles of NH3.
Number of moles of NH3 = 0.03009 mol
Next, calculate the volume of the ammonia solution used.
Volume of ammonia (in L) = 15.0 cm3 * (1 L / 1000 cm3) = 0.015 L
Finally, calculate the molarity of the ammonia:
Molarity of ammonia = Number of moles / Volume in L = 0.03009 mol / 0.015 L = 2.006 M (rounded to three decimal places)
Therefore, the molarity of the ammonia solution is approximately 2.006 M.
2. In the second question, the equation for the reaction between nitric acid and sodium hydroxide is: HNO3 + NaOH -> NaNO3 + H2O.
Based on the balanced equation, the mole ratio between the two reactants is 1:1. This means that 1 mole of HNO3 reacts with 1 mole of NaOH.
First, calculate the number of moles of NaOH used:
Number of moles of NaOH = Molarity * Volume in L = 0.2 M * 0.025 L = 0.005 mol
Since the balanced equation is 1:1 between HNO3 and NaOH, the number of moles of HNO3 required is also 0.005 mol.
Next, calculate the volume of HNO3 needed:
Volume of HNO3 (in L) = Number of moles / Molarity = 0.005 mol / 0.5 M = 0.01 L (or 10 ml)
Therefore, 10 ml of 0.5 M HNO3 is required to neutralize 25 ml of 0.2 M NaOH.
3. In the third question, the equation for the reaction between nitric acid and potassium hydroxide is: HNO3 + KOH -> KNO3 + H2O.
Similar to the previous question, the mole ratio between HNO3 and KOH is 1:1.
First, calculate the number of moles of KOH used:
Number of moles of KOH = Molarity * Volume in L = 0.45 M * 0.02855 L = 0.01246 mol
Since the balanced equation is 1:1 between HNO3 and KOH, the number of moles of HNO3 needed is also 0.01246 mol.
Next, calculate the volume of HNO3 required:
Volume of HNO3 (in L) = Number of moles / Molarity = 0.01246 mol / 0.6 M = 0.0208 L (or 20.8 ml)
Therefore, 20.8 ml of 0.6 M HNO3 is necessary to neutralize 28.55 ml of 0.45 M KOH.