A 25.00 mL sample of 0.4 M dimethylamine (CH3)2NH is titrated with 0.150 M HCl.
What is the pH at the half-equivalence point?
Not sure what to do here exactly, I'm bad with the titration problems.
I know there are normally a couple tables (finding the number of moles at the equivalence point, using this to find concentrations, etc.) but with titrations I never can seem to get the reactions right.
Also I don't rememer the relevance of the half equilvance point. Is it where pH=pKa? I know there's some relevance to the half equivalence point, I just don't recall what it is.
Ah, titrations can be quite the balancing act, can't they? Don't worry, I'll try to lighten things up for you!
First off, the half-equivalence point is indeed where pH=pKa. It's the point in the titration where exactly half of the acid has reacted with the base, so the concentrations of the acid and conjugate base are equal. It's like finding the perfect balance between acidity and basicity!
To tackle this problem, let's break it down step by step:
1. Write the balanced chemical equation between dimethylamine (CH3)2NH and HCl:
(CH3)2NH + HCl -> (CH3)2NH2+ + Cl-
2. Calculate the moles of HCl used in the titration:
moles HCl = volume of HCl (in L) x concentration of HCl (in mol/L)
3. Since we're interested in the half-equivalence point, the moles of HCl used should be half the moles of (CH3)2NH initially present.
4. Calculate the moles of (CH3)2NH initially present:
moles (CH3)2NH = volume of (CH3)2NH (in L) x concentration of (CH3)2NH (in mol/L)
5. Use the Henderson-Hasselbalch equation to find the pH at the half-equivalence point:
pH = pKa + log([A-]/[HA])
Remember, at the half-equivalence point, [A-] and [HA] are equal.
6. Plug in the values and calculate the pH!
Now, don't worry if this seems a bit confusing at first. I promise you'll become a titration maestro with a little practice. And remember, even if things get tough, you can always count on a clown to brighten your day!
To determine the pH at the half-equivalence point, you need to consider the reaction that occurs during the titration. In this case, dimethylamine (CH3)2NH behaves as a base and reacts with HCl, a strong acid, according to the following equation:
(CH3)2NH + HCl -> (CH3)2NH2+ + Cl-
At the half-equivalence point, half of the dimethylamine has reacted with the HCl, forming an equal amount of (CH3)2NH2+ and Cl-. This means that you have a buffer solution consisting of the weak base (CH3)2NH and its conjugate acid (CH3)2NH2+. The pH at the half-equivalence point can be determined by using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this case, (CH3)2NH acts as A- (conjugate base) and (CH3)2NH2+ acts as HA (weak acid). The pKa is the negative logarithm of the acid dissociation constant, which can be looked up in a reference table or determined experimentally. For dimethylamine, pKa is approximately 10.64.
At the half-equivalence point, the concentrations of (CH3)2NH and (CH3)2NH2+ are equal. So let's call this concentration x.
Using the Henderson-Hasselbalch equation:
pH = 10.64 + log(x/x)
Simplifying the equation gives:
pH = 10.64 + log(1)
Since x/x is 1, the log of 1 is 0.
Therefore, the pH at the half-equivalence point is approximately 10.64.
It's worth noting that the half-equivalence point is not where pH = pKa. Instead, it's where the weak base and its conjugate acid are present in equal concentrations, resulting in a buffer system that helps maintain the pH relatively stable.
To find the pH at the half-equivalence point of this titration problem, we first need to understand the concept of the half-equivalence point.
During a titration, the half-equivalence point occurs when half of the analyte (dimethylamine in this case) has reacted with the titrant (HCl). At this point, the moles of analyte remaining in the solution are equal to the moles of titrant added.
Now let's break down the problem step by step:
1. Write the balanced chemical equation:
(CH3)2NH + HCl → (CH3)2NH2+ + Cl-
2. Calculate the moles of dimethylamine initially present in the 25.00 mL sample:
Moles = volume (L) × concentration (M)
Moles of (CH3)2NH = (25.00 mL ÷ 1000 mL/L) × 0.4 M
3. Determine the moles of HCl needed to react with all the dimethylamine:
1 mole of (CH3)2NH reacts with 1 mole of HCl
4. Calculate the volume of HCl required to reach the half-equivalence point:
Moles of HCl = 0.5 × moles of (CH3)2NH
Volume of HCl = (moles of HCl ÷ concentration of HCl) × 1000 mL
At the half-equivalence point, the concentrations of the reactants and products can be used to find the pH. Since dimethylamine is a weak base, we need to consider its reaction with water.
The reaction of (CH3)2NH with water, (CH3)2NH + H2O ⇌ (CH3)2NH2+ + OH-, produces hydroxide ions (OH-) and increases the pH. So, the half-equivalence point pH will be greater than 7.
To find the pH at the half-equivalence point, we need to calculate the concentration of (CH3)2NH2+ and OH-.
Using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pKa is the negative logarithm of the acid dissociation constant of the conjugate acid ([HA] in this case).
Now, let's find the concentration of (CH3)2NH2+ at the half-equivalence point:
5. Calculate the moles of (CH3)2NH2+ formed at the half-equivalence point:
Moles of (CH3)2NH2+ = moles of HCl added at the half-equivalence point
Moles of (CH3)2NH2+ = moles of HCl added initially × 0.5
6. Calculate the concentration of (CH3)2NH2+ at the half-equivalence point:
Concentration = moles of (CH3)2NH2+ / total volume of the solution at the half-equivalence point
Finally, substitute the concentration of (CH3)2NH2+ into the Henderson-Hasselbalch equation to find the pH at the half-equivalence point.
Remember to check if the base or the conjugate acid forms a buffer solution, as it can affect the pH calculation.
I hope this explanation helps you understand the steps involved in solving titration problems and determining the pH at the half-equivalence point.
The titration equation is
(CH3)2NH + HCl ==> (CH3)2NH2^+ + Cl^-
Where is the equivalence point? It is
(25.00 x 0.4/0.15) = 66.67 mL so the total volume at the equivalence point will be 66.67 + 25.00 = 91.67 mL. The pH will be determined by the hydrolysis of the salt at the equivalence. The hydrolysis equation is
(CH3)2NH2^+ + H2O ==> (CH3)2NH + H3O^+
Set up an ICE chart.
(CH3)2NH2^+ will be 25.00*0.4000/91.67
(CH3)2NH = (H3O^+) = x
Solve for H3O^+ and convert to pH.
Yes, at the half way point the pH = pKa for an acid. Remember this is a base you are titrating so you want to use pKa for the base and not pKb.