Aqueous potassium iodate and potassium iodide react in the presence of dilute hydrochloric acid, as shown below. kio3(aq) + 5ki(
1 answer:
You might be interested in
Explanation:
It is known that of
=
.
(a) Relation between and
is as follows.
Putting the values into the above formula as follows.
=
= 3.347
Also, relation between pH and is as follows.
pH =
=
= 3.44
Therefore, pH of the buffer is 3.44.
(b) No. of moles of HCl added =
=
= 0.0116 mol
In the given reaction, will react with
to form
Hence, before the reaction:
No. of moles of =
= 0.15 mol
And, no. of moles of =
= 0.12 mol
On the other hand, after the reaction :
No. of moles of = moles present initially - moles added
= (0.15 - 0.0116) mol
= 0.1384 mol
Moles of = moles present initially + moles added
= (0.12 + 0.0116) mol
= 0.1316 mol
As, =
=
= 3.347
Since, volume is both in numerator and denominator, we can use mol instead of concentration.
As, pH =
= 3.347+ log {0.1384/0.1316}
= 3.369
= 3.37 (approx)
Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.
The question requires us to complete the sentence regarding the preparation of a more dilute NaOH solution (0.100 M, 50.0 mL) from a more concentrated NaOH solution (1.00 M).
Analyzing the blank spaces that we need to fill in the sentence, we can see that we must provide the volume of the more concentrated solution and the volume of water necessary to prepare the solution.
We can use the following equation to calculate the volume of more concentrated solution required:
where C1 is the concentration of the initial solution (C1 = 1.00 M), V1 is the volume required of the inital solution (that we'll calculate), C2 is the concentration of the final solution (C2 = 0.100 M) and V2 is the volume of the final solution (V2 = 50.0 mL).
Applying the values given by the question to the equation above, we'll have:
Thus, we would need 5.00 mL of the more concentrated solution.
Since the volume of the final solution is 50.0 mL and it corresponds to the volume of initial solution + volume of water, we can calculate the volume of water necessary as:
Thus, we would need 45.0 mL of water to prepare the solution.
Therefore, we can complete the sentence given as:
<em>"In order to prepare 50.0 mL of 0.100 M NaOH you will add </em>5.00 mL<em> of 1.00 M NaOH to </em>45.0 mL<em> of water"</em>