Consider the titration of 150.0 mL of 0.100 M HI by 0.250 M NaOH . a. Calculate the pH after 20

step 1 Okay, so this question is actually a titration problem. Basically, ask you to calculate the pH a

Consider the titration of 150.0 mL of 0.100 M HI by 0.250 M...

Key Concepts

Acid-Base Titration

Acid-base titration is a method used to determine the concentration of an acidic or basic solution by gradually adding a titrant of known concentration until the reaction reaches completion. This technique relies on the neutralization reaction between the acid and the base and often involves indicators or instruments to detect the endpoint or equivalence point of the reaction.

Stoichiometry of Neutralization Reactions

The stoichiometry of neutralization reactions involves calculating the number of moles of the reactants—the acid and the base—to determine which reactant is in excess and how much remains after the reaction. By comparing the molar amounts, one can ascertain the progress of the titration and accurately compute the concentration of the remaining species that impact the solution’s pH.

Strong Acid and Strong Base Reaction

In titrations involving strong acids and strong bases, the reaction proceeds to completion because both the acid and the base dissociate completely in aqueous solutions. This characteristic simplifies pH calculations, as the assumption of complete neutralization is justified and the pH can be directly determined from the concentration of the excess hydrogen or hydroxide ions.

Equivalence Point

The equivalence point in an acid-base titration is reached when the amount of titrant added is exactly enough to completely neutralize the analyte. In the case of strong acid/strong base titrations, the equivalence point typically corresponds to a neutral pH of 7, reflecting the complete conversion of acid to water and a salt. Determining the equivalence point is crucial for the accurate analysis of titration curves and for calculating pH at different stages of the titration process.

pH Calculation from Excess Reactant

After the titration has progressed past a certain stage, the pH of the solution is determined by the concentration of the remaining excess reactant—either acid or base. The calculation involves finding the concentration of hydrogen ions or hydroxide ions in the solution post-reaction and then using the pH or pOH formulas to ascertain the final pH of the solution.

Transcript

00:01 Okay, so this question is actually a titration problem.

00:06 Basically, ask you to calculate the ph after mixing acid and base solutions.

00:17 So it says initially we have a solution which is 150 milliliters of the h .i., which is an acid, which is an acid.

00:33 The concentration is 0 .1 more.

00:39 And we are using sodium hydroxide to titrate such solution.

00:49 And the sodium hydroxide has a concentration of 0 .25 more.

00:57 So the first question is that when you add 20 millimeters of the sodium hydroxide into this 150 mililiters, of the h -i solution, what is the ph of the mixture? so in order to solve that, we first need to write down the reaction equation.

01:24 We know that the sodium hydroxide will react with the h -i, and then it forms water and an a -i.

01:43 So first, we need to calculate how much is the h .i.

01:53 The mores of the h .i.

01:59 Equals to the concentration of h .i., which is 0 .1.

02:06 More times the volume, which is 150 militer, which is 10 to minus 3 liters.

02:16 And we can get in the solution, there would be 0 .1 .5 times 10 to minus 2 more of the sodium h .i.

02:41 And how about sodium hydroxide? the more of the sodium hydroxide equals to the concentration, which is 0 .25 more, times of volume, which is 20 milliliter times 10 to minus 3 liter, which equals to 5 times 10 to minus 3, which is 0 .5 times 10 to minus 2 more.

03:17 So you can see here, and we know that one sodium hydroxide we average h .i.

03:25 The ratio is 1 to 1.

03:28 And we know that initially we have 1 .5 times 10 to minus 2 more of the h .i.

03:36 But only 0 .5 times 10 to minus 2.

03:39 So we know h .i is excessive, which means the sodium hydroxide will be totally consumed.

03:52 There will be no sodium hydroxide left in the solution.

03:56 But there would be extra h .i.

03:59 So then we can calculate the first.

04:03 Final moors of the h .i.

04:07 Simply equals to the initial morse of the h .i.

04:11 Minors the consumed h .i.

04:18 The consumed is 1 .5 times 10 to minus 2 minus 0 .5 times minus 2...