A25.00 ml solution of 0.08670 m nai is titrated with 0.05100 m agno 3 . calculate pag + following the addition of the given volumes of agno 3 . the k sp of agi is 8.3 × 10 − 17

a) 13.7

b) 8.04

c) 2.48

Explanation:

In the comments of this question, there's the amount of volume added.

Before we do any calculations of adding, with the first data let's calculate the Equivalence volume. This is the volume where the reaction occurs, and the AgI begins to precipitate.

As the reaction of AgNO3 and AgI has a mole ratio of 1:1, we can say that:

MaVa = MbVb (1)

Where:

MaVa: concentration and volume of NaI

MbVb: concentration and volume of AgNO3

According to this, let's calculate Vb, and this would be the equivalence volume of this titration:

0.08670 * 25 = 0.05100 * Vb

Vb = 42.5 mL

This means that the equivalence point is reached at 42.5 mL, therefore, the first volume of 37.5 mL is before the equivalence point, and the last is after the equivalence point.

Knowing this, let's calculate now the pAg+ at different volumes:

a) When 37.5 mL of AgNO3 is added

In this case, the total volume is 62.5 mL, so let's calculate the original moles of each substance first before the titration:

moles I- = 0.0867 * 0.025 = 2.17x10^-3 moles

moles Ag+ = 0.051 * 0.0375 = 1.91x10^-3 moles

Now, as the reaction taking place is:

AgNO3 + NaI > AgI + NaNO3

we can say that as both have mole ratio of 1:1, we can say that the excedent reactant in these conditions is the NaI, so the remaining moles of NaI (In this case, the I-) are:

moles I- = 2.17x10^-3 - 1.91x10^-3 = 2.6x10^-4 moles

This remaining moles of I-, are the leftovers when the 37.5 mL of Ag+ is added. herefore the concentration of I- has changed:

[I-] = 2.6x10^-4 / 0.0625 = 4.16x10^-3 M

Now, as we want to know the pAg+, we know that the reaction of precipitation taking place is the following:

Ag+ +  I- <> AgI   Ksp = 8.3x10^-17

From this, we can write the expression for Ksp which is:

Ksp = [Ag+] [I-]  (2)

From here, we can solve for [Ag+]:

[Ag+] = Ksp / [I-]  (3)

Now replacing the values of [I-] previously obtained and the Ksp we have:

[Ag+] = 8.3x10^-17 / 4.16x10^-3

[Ag+] = 2x10^-14 M

Finally to calculate the value of pAg+, this is calculated as if you want to calculate the pH of an acid base reaction:

pAg+ = -log[Ag+]  (4)

Replacing we have:

pAg+ = -log(2x10^-14)

pAg+ = 13.7

This is pAg+ after adding 37.5 mL

b) In the equivalence point

In this part, we know that the Ve = 42.5 mL as it was previously calculated. At this point, the moles of both substance are the same, in other words:

moles Ag+ = moles I-

So, the concentrations are given by the general reaction of precipitation, and this can be solve with an ICE chart:

   AgI <> Ag+ + I-   Ksp = 8.3x10^-17

i)                    0      0

c)                    +x     +x

e)                    x       x

Then Ksp = x * x = x²    so:

8.3x10^-17 = x²

x = √8.3x10^-17

x = [Ag+] = [I-] = 9.1x10^-9 M

calculating now pAg+:

pAg+ = -log(9.1x10^-9)

pAg+ = 8.04

c) after adding 47.3 mL of AgNO3

Now that we have passed the equivalence point by almost 5 mL (47.3 - 42.5 = 4.8 mL) we know that AgI is precipitating and the moles of I- should be now less than the moles of Ag+, so:

moles Ag+ = 0.051 * 0.0473 = 2.41x10^-3 moles

This, compared, with the innitial moles of I-, it's now higher, so the remaining moles we are talking about would be the Ag+ so:

moles Ag+ after the addition: 2.41x10^-3 - 2.17x10^-3 = 2.4x10^-4 moles

The rest of the problem, is doing the same as previously done:

[Ag+] = 2.4x10^-4 / 0.0723 = 3.32x10^-3 M

pAg+ = -log(3.32x10^-3)

pAg+ = 2.48

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