Chemistry, 02.10.2019 13:30, murtaghliam1

Starting with 0.250l of a buffer solution containing 0.250 m benzoic acid (c6h5cooh) and 0.20 m sodium benzoate (c6h5coona), what will the ph of the solution be after the addition of 25.0 ml of 0.100m hcl? (ka (c6h5cooh) = 6.5 x 10-5)

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answered: piggygirl211

New pH will be 4.06 .

Explanation: The problem could easily be solved using Handerson equation. Handerson equaton is used to calculate the pH of buffer solution.

$pH=pK_a+log(\frac{base}{acid})$

For the given problem, the buffer solution is a mixture of a weak acid(benzoic acid) and a salt of it, known as conjugate base(sodium benzoate).

When a strong acid is added to the buffer solution then it reacts with the base(benzoate ion) present in the buffer solution and produce a the weak acid(benzoic acid).

The reaction could be shown as:

$C_6H_5COO^-(aq)+H^+(aq)\rightleftharpoons C_6H_5COOH(aq)$

First of all we calculate the initial moles of acid and base originally present in the buffer solution and for this the volume is multiplied by the molarity.

initial moles of benzoic acid = $0.250L(\frac{0.250mol}{1L})$

= 0.0625 moles

initial moles of benzoate ion = $0.250L(\frac{0.20mol}{1L})$

= 0.0500 moles

moles of HCl or H^+ added to the buffer = $25.0mL(\frac{1L}{1000mL})(\frac{0.100mol}{1L})$

= 0.0025 moles

From the equation we have written above, HCl and benzoate ion react in 1:1 mol ratio. As HCl is limiting reactant, 0.0025 moles of it will react with exactly 0.0025 moles of benzoate ion and form 0.0025 moles of benzoic acid.

So, the moles of benzoic acid after addtion of HCl = 0.0625 + 0.0025 = 0.065 moles

moles of benzoate ion after addition of HCl = 0.0500 - 0.0025 = 0.0475 moles

Total volume of the solution = 0.250 L + 0.025 L = 0.275 L

concentration of benzoic acid = $\frac{0.065mol}{0.275L}$

= 0.236M

concentration of benzoate ion = $\frac{0.0475mol}{0.275L}$

= 0.173M

Pka is calcuulated from the given Ka value as:

pK_a=-logK_a

pk_a=-log(6.5*10^-^5)

pK_a = 4.19

Let's plug in the values in the Handerson equation:

$pH=4.19+log(\frac{0.173}{0.236})$

pH = 4.19 - 0.13

pH = 4.06

So, the pH of the solution after an addition of HCl will be 4.06 .

answered: Guest

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is $1s^{2}2s^{2}2p^{6}3s^{1}$ and electronic configuration of phosphorous is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$ .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

$3na + p \rightarrow na_{3}p$

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.