Consider a 10.0 % (by mass) solution of hypochlorous acid. Assume the density of the solution to

step 1 Okay, so we have a weak acid, and we know we have 10 % of it. So I'm going to assume that I have

Consider a 10.0 % (by mass) solution of hypochlorous acid....

Key Concepts

Titration Equivalence Point

In acid-base titration, the equivalence point is reached when the amount of titrant added is stoichiometrically equivalent to the amount of the substance being titrated. At this point, all of the original weak acid has been converted into its conjugate base. Calculating the pH at the equivalence point involves considerations of the properties of the conjugate base, such as its ability to hydrolyze in water, which affects the overall pH of the solution. Understanding the equivalence point is essential for determining the complete titration curve of a reacting system.

Buffer Solutions and the Henderson–Hasselbalch Equation

A buffer solution contains significant amounts of both a weak acid and its conjugate base, which allows the solution to resist changes in pH upon the addition of small amounts of acid or base. The Henderson–Hasselbalch equation is used to relate the pH of a buffer solution to the pKa (negative logarithm of the acid dissociation constant) and the ratio of the concentrations of the conjugate base to the weak acid. This equation simplifies the calculation of pH, especially at the half-equivalence point during titration where the concentrations of the acid and conjugate base are equal.

Acid-Base Equilibrium

Acid-base equilibrium involves the dissociation of an acid into its ions in aqueous solution, described by its acid dissociation constant (Ka). The extent of this dissociation determines the concentration of hydrogen ions in the solution, which directly affects the pH. Understanding this equilibrium is key to calculating the pH of weak acid solutions by setting up the equilibrium expression and solving for the hydrogen ion concentration.

Density and Molarity Calculations

The density of a solution is a measure of its mass per unit volume and is used to convert between the mass of the solution and its volume. This conversion is crucial when determining the molarity of a solution from a percent by mass concentration, as it allows for the calculation of moles of solute per liter. Molarity is a central concept in stoichiometry and acid-base titration analyses because it directly relates to the number of reacting particles in a given volume.

Percent by Mass Concentration

This concept refers to the mass of solute present in a given mass of solution expressed as a percentage. It allows one to determine how many grams of the solute are present per 100 grams of the entire solution. When combined with the density of the solution, it becomes possible to find the concentration in terms of moles per liter, which is essential for calculating reaction stoichiometry and pH in titration problems.

Transcript

00:01 Okay, so we have a weak acid, and we know we have 10 % of it.

00:05 So i'm going to assume that i have 10 grams of my hocl for every 100 grams of solution.

00:13 That would give me 10%.

00:15 So we'll go ahead and start by changing grams to moles here, using the molar mass, which is 52 .46.

00:25 So we'll get 0 .191 moles of our hocl.

00:33 They said the density of our solution was 1, so that gives us 100 mil liters of solution, or 0 .1 liter.

00:42 So if we take our moles and divide it by 0 .1 liter, we'll see that our molarity is 1 .91 molar h .o .c.

00:54 So before we titrate anything, what we have is 1 .91 molar of h .o .c .l.

01:00 To find the ph of a weak acid, we're going to go ahead and make an ice plate.

01:05 Box.

01:06 So we'll write the equation for the ionization of our weak acid.

01:12 And we'll keep track of what's happening here.

01:15 To start with, i had 1 .91 molar, and actually none of these minus x plus x, 1 .91 minus x.

01:28 So we'll look up the ka for this, and we'll get that it's 2 .8 times 10 of the negative of 8, so that will be equal to x squared over 1 .91 minus x.

01:45 Okay, and we'll make the typical assumption here.

01:47 We're going to assume, since it's a weak acid, that x is much, much smaller than the original concentration of 1 .91 molar.

01:56 Then if we do that, we can get rid of this x, and that will simplify our math.

02:01 So if we solve for x here, we'll get 2 .3 times 10 to negative 4.

02:09 Molar, and this is the molarity of our hydrogen ion.

02:14 So if we take minus the log of that, we'll get the ph.

02:17 So ph is minus the log of that number, so ph is 3 .64.

02:29 They've asked us for the ph at the half equivalence point.

02:33 So we know, i hope, that the half equivalence point, if we look at this equation for buffers, h plus is ka times moles, acid over moles of base.

02:53 If you're exactly at the half equivalence point, half of the moles of acid have been converted into moles of base.

03:00 So these two things are equal, so that will just be one.

03:04 So the concentration of h plus simply equals ka, which we said was 2 .8 times 10 and negative 8.

03:14 So there's your molarity of your h plus.

03:17 So again, if we take minus the log of that, we'll get the ph at the half equivalence point.

03:24 That'll be 7 .55.

03:27 So now at the equivalence point, we know our weak acid has reacted with strong base to make a weak base in water...

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