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3 years ago

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What is the ph of a buffer prepared by adding 0.809 mol of the weak acid ha to 0.608 mol of naa in 2.00 l of solution? The disso

ciation constant ka of ha is 5.66×10−7.

1 answer:

Anarel [89]3 years ago

5 0

The pH of the buffer is 6.1236.

Explanation:

The strength of any acid solution can be obtained by determining their pH. Even the buffer solution strength of the weak acid can be determined using pH. As the dissociation constant is given, we can determine the pKa value as the negative log of dissociation constant value.

$pKa=-log[H] = - log [ 5.66 * 10^{-7}]\\ \\pka = 7 - log (5.66)=7-0.753=6.247\\\\pka = 6.247$

The pH of the buffer can be known as

$pH = pK_{a} + log[\frac{[A-]}{[HA]}}]$

The concentration of $[A^{-}] = Moles of [A]/Total volume = 0.608/2 = 0.304 M\\$

Similarly, the concentration of [HA] = $\frac{Moles of HA}{Total volume} = \frac{0.809}{2} = 0.404$

Then the pH of the buffer will be

pH = 6.247 + log [ 0.304/0.404]

$pH = 6.247 + log 0.304 - log 0.404=6.247-0.517+0.3936=6.1236$

So, the pH of the buffer is 6.1236.

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3 years ago

Calculate Ho298 for the process

Inga [223]

Explanation:

As per the Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

Hence, according to this law the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

$Sb + \frac{3}{2}Cl_2 \rightarrow SbCl_{3}$ $\Delta H^0_1 = -314 kJ$ ..........(1)

$SbCl_{3} + Cl_2 \rightarrow SbCl_{5}$ $\Delta H^0_2 = -80kJ$ ..............(2)

The final reaction is as follows:

$Sb + \frac{5}{2}Cl_{2} \rightarrow SbCl_{5}$ $\Delta H^0_3 = ?$ .............(3)

Therefore, adding (1) and (2) we get the final equation (3) and value of $\Delta H^{0}_{3}$ at 298 K will be as follows.

$\Delta H^{0}_{3}$ = $\Delta H^{0}_{1}$ + $\Delta H^{0}_{2}$

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3 years ago

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2 years ago

You need to use a dilute hydrochloric acid solution in an experiment. However, the only bottle of hydrochloric acid in your lab’

mr_godi [17]

The pH of the diluted HCl solution is 1.3.

Explanation:

Given:

The concentrated HCl solution of 8.0 M. The 1.5 mL of 8.0 M HCl is diluted with water to 250 mL volume.

To find:

The pH of the diluted HCl solution.

Solution

The concentration of the HCl solution before dilution = $M_1=8.0 M$
The volume of the HCl solution taken for dilution = $V_1=1.5 ml$
The concentration of the HCl solution after dilution = $M_2=?$
The volume of the HCl solution after dilution = $V_2=250 mL$

Using the Dilution equation:

$M_1V_1=M_2V_2\\8.0M\times 1.5ml=M_2\times 250 mL\\M_2=\frac{8.0M\times 1.5ml}{250 mL}=0.048 M$

The concentration of diluted HCl solution = 0.048 M

$HCl(aq)\rightarrow H^+(aq)+Cl^-(aq)$

In the 1 M solution of HCl, there are 1 M of hydrogen ion, then the concentration of hydrogen ions in 0.048 M of HCl will be:

$[H^+]=1\times 0.048M=0.048 M$

The pH of the diluted HCl solution :

$pH=-\log [H^+]\\=-\log [0.048M]=1.18 \approx 1.3$

The pH of the diluted HCl solution is 1.3.

Learn more about the dilution equation here:

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2 years ago