Question

Consider the titration of a 27.5 −mL sample of 0.120 MRbOH with 0.105 M HCl. Determine each of the following.

1) the initial pH

2) the volume of added acid required to reach the equivalence point

3) the pH at 5.9 mL of added acid

4) the pH at the equivalence point

5) the pH after adding 5.7 mL of acid beyond the equivalence point

Answer 1

1) the question says titration of a strong base (RbOH) with a strong acid (HCl).

As we
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

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

[H+] (0.120M) = 1x10^-14

which gives us [H+] = 8.3333 x 10^-14 M

also we know that pH= -log(H+)

So pH = -log (8.3333 x 10^-14M) = 13.07

2) since at the equivalent point in a titration,

moles of acid = moles of base.

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

n= 0.120 M OH- x 0.0275 L = 0.0033 moles OH- initially present.

So at equilibrium, to neutralize 0.0033 moles base requires0.0033 moles  acid added.

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

0.0033 moles H+ = (0.105M H+) * V, so V =0.031428L or 31.428 mL

3) the pH at 5.9 mL of added acid


we first find the of moles of acid we are adding :

no of moles of acid = C x V;

n = (0.105 M H+)(0.0059L) => 0.0006195 moles H+ added.

now the remainig base left =moles of total base - moles of acid added :

0.0033 moles - 0.0006195 moles = 0.0026805 moles OH- left.

the total volume becomes = initial volume + volume of acid added :

0.0275 L + 0.0059 L = 0.0334 L

as Concentration = no of moles /Volume of solution ;

[OH-] = 0.0026805 moles OH-/0.0334 L

[OH-] = 0.0825 M and substituting this into our pH equation

as [H+][OH-] = Kw = 1 x 10^-14

[H+] = 1 * 10^-14 / 0.0825 = 12.1212 * 10^-14

pH = -log (* 10^-14 ) = 12.9164

4)

At the equivalence point: moles base = moles of acid there fore

[H+] = 1 x10^-7, so pH = 7.

5)
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 = (0.105 M HCl) x (0.0057L) = 0.0005985 moles H+

total volume = initial volume + volume of HCl at equivalence + volume at the added hcl after equvalence

= 27.5 + 31.428 +  5.7 = 64.628 ml

our concentration of [H+] then = (0.0005985 moles) / (0.064628 L) = 0.00926 M H+

Then:

pH = -log ( 0.00926 M H+) = 2.033

Our pH is now 2.033 .