Consider the titration of 50.0 mL of 0.116 M NaOH with 0.0750 M HCl. Calculate the pH after the

Consider the titration of 50.0 mL of 0.116 M NaOH with...

Key Concepts

Acid-Base Neutralization

This concept involves the reaction between an acid and a base that results in the formation of water and a salt. In a titration involving strong acid and strong base, the reaction proceeds quantitatively according to a 1:1 molar ratio. Understanding this reaction is critical in determining which reactant is in excess after mixing the two solutions.

Molarity and Volume Calculations

Molarity, defined as moles of solute per liter of solution, is used together with the volume of the solution to calculate the number of moles present. This calculation is essential in titration problems where the initial amounts of the acid and base are found by multiplying their respective concentrations by the volumes used.

Stoichiometry and Limiting Reagent Determination

Stoichiometry involves using the mole ratio from the balanced chemical equation to determine the amounts of reactants that will react and which one is the limiting reagent. In titrations, comparing the moles of acid added with the moles of base originally present allows the identification of the excess species after the reaction.

pH Calculation in Strong Acid-Base Titrations

In titrations with strong acids and bases, the remaining concentration of either the excess acid or base determines the pH of the solution. When an excess of strong base remains, pOH is calculated from its concentration and then converted to pH, while an excess of strong acid directly contributes to a lower pH value.

Dilution Effects in Titrations

As the titrant is added, the total volume of the solution changes, affecting the concentration of the remaining reactive species. Dilution effects must be taken into account when calculating the final concentration that is used to determine the pH of the solution at various points during the titration.

Transcript

00:01 Here, part a of the problem, the very first thing that we will need to do is we will have to find the initial moles of n -a -o -h.

00:08 And how we do that is simple dimensional analysis.

00:12 We're going to take the volume of n -a -o -h that we are adding, and we're going to multiply it by one liter is every 1 ,000 milliliters, multiplied by the moles of h -cl per every 1 -liter of solution.

00:27 And when we do this, we will get 0 .0058 moles.

00:32 Now we need the volume of hcl needed in our solution.

00:37 So we're going to take the moles that we found before and multiply it by our 1 -to -1 molar ratio of our reaction.

00:46 And multiply it for every one liter of a solution, we have 0 .075 moles of hcl.

00:53 And we will get 77 .3 milliliters.

00:59 And now we have to find the moles of hcl that we have consumed.

01:03 So we're going to take 5 milliliters of hcl.

01:07 Multiply, by every 1 liter is 1 ,000 milliliters, times 0 .07 moles per every 1 liter of solution, and then again multiplied by our 1 to 1 ratio of our reaction.

01:20 And now we need to find the sodium hydroxide that is remaining.

01:25 So we take our initial moles that we found, subtract the moles that we just found that is consumed in our reaction, and we will get our remaining.

01:34 Moles.

01:35 0 .00542 moles.

01:40 And now we have to find the molarity of sodium hydroxide in this first part.

01:46 So we'll take the moles that we just calculated, divided by our total volume, 55 milliliters, multiply it by 1 ,000 for every 1 liter to get this into liters, and we will get our concentration or molarity .0985 molar.

02:04 And now we're one step closer to 5 .5 molar.

02:06 Finding our ph.

02:08 So our ph in our first part is 14 plus the log of our hydroxide concentration.

02:14 And that's simple...

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