A 20.00 mL sample of each of the following acid solutions...

$ \mathrm{~A} 20.00 \mathrm{~mL}$ sample of each of the following acid solutions is to be titrated to the equivalence point using a 0.120 M NaOH solution. Determine the number of milliliters of NaOH solution that will be needed for each acid sample. a. $0.200 \mathrm{M} \mathrm{HClO}_4$ b. $0.125 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$ c. $0.150 \mathrm{M} \mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_6$ d. $0.120 \mathrm{~mol} \mathrm{H}_3 \mathrm{PO}_4$ in 500.0 mL of solution e. 6.25 g of $\mathrm{H}_2 \mathrm{SO}_4$ in 250.0 mL of solution f. $0.500 \mathrm{~mol} \mathrm{HClO}_3$ in 1.00 L of solution

$ \mathrm{~A} 20.00 \mathrm{~mL}$ sample of each of the following acid solutions is to be titrated to the equivalence point using a 0.120 M NaOH solution. Determine the number of milliliters of NaOH solution that will be needed for each acid sample.
a. $0.200 \mathrm{M} \mathrm{HClO}_4$
b. $0.125 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$
c. $0.150 \mathrm{M} \mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_6$
d. $0.120 \mathrm{~mol} \mathrm{H}_3 \mathrm{PO}_4$ in 500.0 mL of solution
e. 6.25 g of $\mathrm{H}_2 \mathrm{SO}_4$ in 250.0 mL of solution
f. $0.500 \mathrm{~mol} \mathrm{HClO}_3$ in 1.00 L of solution

Key Concepts

Dilution and Concentration Calculations

In cases where a solute is dissolved in a larger volume than the given sample, dilution effects must be considered to determine the actual concentration of the analyte in the titrated portion. This involves using the relationship between initial and final volumes to calculate the effective concentration used in titration.

Polyprotic Acids

Polyprotic acids are acids that can donate more than one proton (H+ ion) per molecule in a reaction. In titrations involving polyprotic acids, the stoichiometric relationships become more complex because each acid molecule contributes multiple equivalents of H+, which must be taken into account when calculating the titration volumes.

Mass-to-Mole Conversion

When the amount of an acid or any reactant is given in mass, it is necessary to convert this mass into moles using the molar mass of the substance. This conversion is critical for stoichiometric calculations in titrations and other chemical reactions, ensuring that the quantities of reactants are accurately compared.

Acid-Base Titration and Equivalence Point

Acid-base titration is an analytical technique used to determine the concentration of an acid or base in a solution by reacting it with a solution of known concentration. The equivalence point is the stage in the titration at which the number of moles of titrant exactly neutralizes the number of moles of the analyte, taking into account the stoichiometry of the reaction.

Molarity

Molarity is a measure of concentration expressed as the number of moles of solute per liter of solution. It serves as a fundamental concept in solution chemistry, allowing for the calculation of how many moles of a reactant are present in a given volume of a solution.

Titration Stoichiometry

Titration stoichiometry involves the use of balanced chemical equations to relate the moles of acid and base involved in a titration. This concept is key for determining the required volume of titrant needed to reach the equivalence point, particularly when acids or bases undergo reactions with different proton donation or acceptance ratios.

Transcript

00:01 This is a pretty easy problem, but there are six of them to do, so let's get right to it.

00:05 We are asked to calculate the volume of acid required to react, and this is the same for everything, with 50 milliliters of 0 .10 -molar sodium hydroxide.

00:18 So our first order of business will be to calculate what we have for moles of sodium hydroxide.

00:25 Let's do that by taking our milliliters, converting it to liters.

00:33 This is n -a -o -h times 0 .1 -100, yep.

00:43 And that's going to equal 5 .00 times 10 to the minus 3 moles of n -a -o -h for each one of these that we're doing.

00:55 And the acids.

01:00 Okay, let's start with number one.

01:03 I got to get to my table here so i know what we're doing.

01:08 Whoopsie.

01:11 Sorry about this.

01:12 I'm trying to do this left -handed, and i am not ambidextrous.

01:17 Our first scenario, we have 0 .1 -00 molar hcl.

01:26 I'm not going to use states or anything, but i'm just going to write these out super quick.

01:40 We can see that these are all one -to -one ratios, so i'll probably do these in a little bit less detail, but i'm going to go ahead and set this one up as 5 .0 times 10 to the minus 3, moles of naoh.

02:03 We've got a 1 to 1 mole ratio and we were given 0 .10 moles per liter for our acid concentration.

02:25 So you can see on this that it's going to be did i do that first one right? 5 times 10 to the minus 2 and to the right number of figs that'll be 50 .0 mill liter.

02:51 Of hcl, and let's even be more specific, milliliters of 0 .10 -0 molar hcl.

03:09 Next.

03:13 Our next substance for b or number two, we are given, we're going to have n -a -o -h, we're given sulfuric gas or sulfurous acid.

03:32 I should go find out which one this is.

03:37 0 .100 molar sulfuric acid.

03:49 And for three, we will be given 0 .2 -00 molar phosphoric acid.

04:03 And for number four, we are given 0 .150 molar nitric acid.

04:12 Okay.

04:17 So there, we've got a few of these that we can do now.

04:20 And let's get those started.

04:22 So with my sulfurus, i've got n -a -2 -s -o -3 plus.

04:35 And this will not be balanced, so i've got a two there and a two there.