Consider the titration of a 28.0 -mL sample of 0.125 M RbOH with 0.110 M HCl. Determine each of the following.?

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a) the initial pH b) the volume of added acid required to reach the equivalence point c) the pH at 5.1 mL of added acid d) the pH at the equivalence point e) the pH after adding ...show more
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  • Cian answered 5 years ago
This problem should be worth way more than 10 points, but here goes:

This is a titration of a strong base (RbOH) with a strong acid (HCl).

PART A.

We know that: pH = -log [H+]

So we need to find [H+] to get the pH of the solution. But initially, we only have [OH-] in solution.

However, because of the autoionization of water, we know that:
[H+][OH-] = Kw = 1 x 10^-14
So now we have our link between [H+] and [OH-].

Let's plug in what we know.

Initial [OH-] = 0.125 M OH- , therefore:

[H+] (0.125M) = 1x10^-14

which gives us [H+] = 8.00 x 10^-14 M
(Because Kw is a constant, I'm sticking with the 3 sig figs that 0.125 M gives us for our answer) This value should make sense because with such a high [OH-], there should be very little [H+].

So pH = -log (8.00 x 10^-14M) = 13.097 (answer has same # of decimal places as input value has sig figs - 3 sig figs = 3 decimal places after decimal point).

PART B.

at the equivalent point in a titration, moles of acid = moles of base.

moles (n) = C x V (C is concentration and V is volume in Liters).

n= 0.125 M OH- x 0.0280 L = 0.00350 moles OH- initially present.

So at equilibrium, to neutralize 0.00350 moles base requires 0.00350 moles acid added.

To find the Volume of acid, use the moles equation again using the concentration of acid used:

0.00350 moles H+ = (0.110M H+) x V, so V=0.0318L or 31.8 mL

PART C

In order to reach the equivalence point, we just found out we need to add 31.8 mL of the 0.110M HCl. Here, we've only added 5.1 mL - so we have not reached the equivalence point yet - but we have neutralized SOME of the moles of base initially present.

lets first find the # of moles of acid we are adding here:

n = C x V; n = (0.110M H+)(0.0051L) ===> 0.00056 moles H+ added.

Remember our initial moles of base was 0.00350 moles. By adding 0.0056 moles of acid, we neutralize the same moles of base. To find what is left over subtract what we added from what was initially present:

0.00350 moles - 0.00056 moles = 0.00294 moles OH- left.

In order to find our new concentration, we need the total volume now:

0.0280L + 0.0051L = 0.0331L

C = n/V; [OH-] = 0.00294 moles OH-/ 0.0331L

[OH-] = 0.0889 M and substituting this into our pH equation PROPERLY:

pH = -log ( 1x10^-14 / 0.00889 M) = 12.948

This is the pH after adding 5.1 mL of our 0.110 M HCl.

PART D.

At the equivalence point: moles base = moles of acid. we have no moles of either left over , but because of the autoionization of water, [H+] = 1 x10^-7, so pH = 7.

PART E.

To find the pH after the equivalence point, we need the [H+] after the equivalence point, so we need to know how many moles of acid are present and our total volume of solution to find this concentration value.

moles = (0.110 M HCl) x (0.0042L) = 0.00046 moles H+

our volume started at 28.0mL, then we added 31.8 mL to get to the eq. point (part b above). Now we are adding 4.2 mL beyond that - giving us a total of 64.0 mL or 0.0640L.

our [H+] then = (0.00046 moles) / (0.0640L) = 0.0072 M H+

Then:

pH = -log (0.0072 M H+) = 2.14

Our pH is now 2.14.

Source:

My M.S. in chemistry.

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