Consider the titration of 30.0 mL of 0.050 M NH3 ...

Question

Consider the titration of 30.0 mL of 0.050 M  NH3 with 0.025 M  HCl. Calculate the pH after the following volumes of titrant have been added.20.0 mL

Jules Bruno · Accepted Answer

We’re being asked to calculate the pH of a solution after the addition of 20.0 mL 0.0250 M HCl solution. The solution is made up of 30.0 mL of 0.050 M ammonia (NH3).HCl (strong acid) will react with NH3 (base).▪ NH3 is a base and based on Bronsted-Lowry definition it is a proton acceptor.▪ HCl is an acid and based on Bronsted-Lowry definition it is a proton donor.Reaction:NH3(aq) + HCl(aq) → NH4+(aq) + Cl-(aq)We will calculate the pH of the solution using the following steps:[readmore]Step 1: Calculate the initial amount (in moles) of each species.Step 2: Construct an ICF chart to determine the final amount of each species.            Recall that when we’re using moles, we’re going to use an ICF ChartStep 3: Calculate the pH of the solution.Step 1: Calculate the initial amount (in moles) of each species:Given:• 30.0 mL of 0.050 M NH3• 20.0 mL of 0.0250 M HClmolarity (M)=molLmoles NH3=0.050mol NH3L 30.0 mL x 10-3 L1 mLmoles NH3 = 0.0015 mol NH3moles HCl=0.0250mol HClL 20.0 mL x 10-3 L1 mLmoles HCl = 5 x 10-4 mol HClStep 2: Construct an ICF chart to determine the final amount of each species.▪ HCl has the least amount so it will be totally consumed in the reaction▪ lost some reactant → NH3 reacted with HCl            ▪ 1:1 mole ratio for the acid and base reaction, same amount of NH3 will be consumed▪ gained some products → NH4+▪ 1:1 mole ratio for NH3 and NH4+, same amount of NH4+ will be produced▪ Cl- is from a strong acid → neutral ion → will not contribute to the pH of the solutionAfter the reaction is complete the solution contains:• NH4+ is the conjugate acid of NH3 → weak acid▪ when NH4+ is in aqueous solution, it reacts with water:  NH4+(aq)     +      H2O(l)      →      NH3(aq)     +       H3O+(aq)(weak acid)             (base)           (conjugate base)    (conjugate acid)• 5 x 10-4 mol NH4+ → weak acid• 0.001 mol NH3 → conjugate base• weak acid + conjugate base = bufferStep 3: Calculate the pH of the solution Since we have a buffer we can use the Henderson-Hasselbalch Equation to calculate for pH.• 5 x 10-4 or 0.0005 mol NH4+• 0.001 mol NH3pH=pKa+logconjugate baseweak acidKb for NH3 = 1.8x10-5 (refer to textbooks/online)We need pKa and Ka for the Henderson-Hasselbalch equation. Ka and Kb are connected by the following equation:Kw=Ka·KbWhere Kw is the autoionization constant of water:KwKb=Ka·KbKbKa=KwKbAssuming the reaction is at 25°C:  Kw = 1.0x10-14Ka=1.0x10-141.8x10-5Ka = 5.55x10-10pKa can be calculated from Ka using the following equation:pKa=-log KaSubstitute in the Henderson-Hasselbalch Equation:pH=-log Ka+log conjugate base weak acidpH=-log 5.55x10-10+log 0.001 mol0.0005 molpH=9.256+log (0.301)pH=9.256+(-0.521)pH=9.256-0.521pH = 8.74The pH of the solution after adding HCl is 8.74.

Problem: Consider the titration of 30.0 mL of 0.050 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added.20.0 mL

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Problem: Consider the titration of 30.0 mL of 0.050 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added.20.0 mL

FREE Expert Solution

Problem Details

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