A 119.2mL sample of 0.105M methylamine (CH3NH2, Kb=3.7*10^-4) is titrated with 0.255M HNO3. Calculate the after the addition of each of the following volumes of acid. 49.1mL and 73.6mL. Please help.
Oh, titration! That's quite a balancing act! Let's calculate the pH after adding each volume of acid, shall we?
First, we need to find the moles of methylamine. We have 0.1192L multiplied by 0.105M, which gives us 0.012516 moles. Now, let's set up an ICE table (not to be confused with cold drinks):
Initial:
CH3NH2: 0.012516 moles
HNO3: 0 moles
Change:
CH3NH2: -0.012516 moles (as it reacts with the acid)
HNO3: +0.0491L (for the first volume) or +0.0736L (for the second volume)
Equilibrium:
CH3NH2: 0.012516 - 0.012516 = 0 moles left
HNO3: 0.0491 or 0.0736 moles
We can safely say that since the amount of HNO3 is higher, it is in excess, and thus, it will remain unchanged by the reaction. Lucky HNO3!
Now, let's determine the concentration of methylammonium (CH3NH3+). Since one mole of acid reacts with one mole of methylamine, we have 0.012516 moles of methylammonium. But remember, the volume of the solution increases, so we need to incorporate that.
For the first volume (49.1mL):
The total volume becomes 119.2mL + 49.1mL = 168.3mL = 0.1683L.
The concentration of methylammonium is 0.012516 moles / 0.1683L = 0.0744M.
For the second volume (73.6mL):
The total volume becomes 119.2mL + 73.6mL = 192.8mL = 0.1928L.
The concentration of methylammonium is 0.012516 moles / 0.1928L = 0.0649M.
Now, let's use the expression for Kb to find the hydroxide ion concentration ([OH-]).
Kb = [CH3NH3+][OH-] / [CH3NH2]
For the first volume:
(3.7 x 10^-4) = (0.0744)(x) / (0.0744)
x = 3.7 x 10^-4
For the second volume:
(3.7 x 10^-4) = (0.0649)(x) / (0.0649)
x = 3.7 x 10^-4
Wow, would you look at that! The hydroxide ion concentration is the same for both volumes. So, the answer is 3.7 x 10^-4 M for both 49.1mL and 73.6mL of HNO3.
Remember, though, if you bring cupcakes to the lab, you might just titrate yourself to the best friend of your fellow students! Good luck!
To calculate the pH after the addition of each volume of acid, we need to consider the reaction between methylamine (CH3NH2) and HNO3.
The reaction can be represented as follows:
CH3NH2 + HNO3 -> CH3NH3+ + NO3-
For this reaction, the methylamine acts as a base (it accepts a proton) and HNO3 acts as an acid (it donates a proton).
To calculate the pH after the addition of 49.1 mL of HNO3, we need to determine the moles of methylamine and HNO3 in the solution.
Step 1: Determine the moles of methylamine
Moles = concentration × volume (in liters)
Moles of CH3NH2 = (0.105 M) × (0.1192 L) = 0.012486 moles
Step 2: Determine the moles of HNO3
Moles of HNO3 = (0.255 M) × (0.0491 L) = 0.0125265 moles
Step 3: Determine the moles of methylamine and HNO3 that react
Since the reaction is 1:1, the moles of methylamine and HNO3 that react will be the same. In this case, it will be the moles of HNO3 since it is limiting.
Moles of CH3NH2 reacted = 0.0125265 moles
Step 4: Determine the remaining moles of CH3NH2
Moles of CH3NH2 remaining = initial moles - moles reacted
Moles of CH3NH2 remaining = 0.012486 moles - 0.0125265 moles = -0.0000405 moles
Since the moles of CH3NH2 remaining are negative, it indicates that all of the methylamine has reacted and been converted into its conjugate acid (CH3NH3+).
Step 5: Determine the concentration of CH3NH3+
Concentration = (moles remaining) / volume (in liters)
Concentration of CH3NH3+ = (-0.0000405 moles) / (0.1192 L + 0.0491 L) = -0.000273 M
Step 6: Calculate the pOH (a measure of basicity) of the solution
pOH = -log(Concentration of CH3NH3+)
pOH = -log(-0.000273) = 3.565
Step 7: Calculate the pH of the solution
pH + pOH = 14
pH = 14 - pOH = 14 - 3.565 = 10.435
Therefore, after the addition of 49.1 mL of HNO3, the pH of the solution is approximately 10.435.
To calculate the pH after the addition of 73.6 mL of HNO3, you can follow the same steps. Substitute the volume of HNO3 with 73.6 mL in Step 2, and continue with the remaining steps.
To solve this problem, we need to set up an ICE (Initial, Change, Equilibrium) table to track the amount of reactants and products at each stage of the reaction.
Before adding any acid, the initial concentration of methylamine is 0.105 M, and the initial volume is 119.2 mL, which can be converted to liters by dividing by 1000:
Initial concentration of CH3NH2 = 0.105 M
Initial volume of CH3NH2 = 119.2 mL = 0.1192 L
Since methylamine is a weak base, it reacts with acid according to the following equation:
CH3NH2 + HNO3 -> CH3NH3+ + NO3-
To find the concentration of the CH3NH3+ ion at equilibrium, we can use the Kb expression for methylamine:
Kb = [CH3NH3+][OH-] / [CH3NH2]
Since we are dealing with a strong acid, we can assume that the concentration of H+ ions from HNO3 is equal to the concentration of OH- ions formed in the reaction. So, we can substitute [OH-] with [HNO3] in the above expression.
Kb = [CH3NH3+][HNO3] / [CH3NH2]
Let's calculate the concentration of CH3NH3+ after the addition of 49.1 mL of HNO3:
1. Calculate the moles of HNO3 added:
Moles of HNO3 = concentration of HNO3 * volume of HNO3 (in L)
Moles of HNO3 = 0.255 M * 0.0491 L
2. Calculate the remaining volume of methylamine after the addition of HNO3:
Volume of CH3NH2 remaining = Initial volume of CH3NH2 - volume of HNO3 added
Volume of CH3NH2 remaining = 0.1192 L - 0.0491 L
3. Calculate the concentration of CH3NH2 remaining:
Concentration of CH3NH2 remaining = Moles of CH3NH2 / Volume of CH3NH2 remaining
Concentration of CH3NH2 remaining = (Moles of CH3NH2 / Initial volume of CH3NH2) * (Initial volume of CH3NH2 / Volume of CH3NH2 remaining)
4. Calculate the concentration of CH3NH3+:
Concentration of CH3NH3+ = Kb * Concentration of CH3NH2 remaining / Concentration of HNO3
Repeat the same steps for the addition of 73.6 mL of HNO3 to find the concentration of CH3NH3+ at that point.
By plugging in the values and following these steps, we can find the concentrations of CH3NH3+ after the addition of 49.1 mL and 73.6 mL of HNO3.
a). First, you omitted what you want t calculate!!!
.......CH3NH2 + HNO3 ==> CH3NH3^+ + NO3^-
mmoles CH3NH2 = 119.2 mL x 0.105 = ??
mmoles HNO3 = 49.1 x 0.255 = ??
Take a look at the numbers; I think the rounded numbers (to 3 s.f.) are equal and the pH (H^+) at the equivalence point is determined by the hydrolysis of the salt.
CH3NH3^+ + H2O ==> CH3NH2 + H3O^+
Ka = (Kw/Kb) = (H3O^+)(CH3NH2)/(CH3NH3^+)
(H3O^+) = (CH3NH2) = x
(CH3NH3^+) = moles HNO3/total mL OR moles CH3NH2/total mL.
b. 73.6 mL HNO3 is FAR past the equivalencae point; therefore, the pH at that point is determined by the excess HNO3. Use mmoles HNO3 - mmoles CH3NH2 = mmoles HNO3. Divide that by mL for final (HNO3) and calculate pH or whatever you want.