Chemistry, 26.03.2020 20:36 gennhill14

A chemist titrates 100.0mL of a 0.3444M hydrocyanic acid HCN solution with 0.8414M KOH solution at 25°C . Calculate the pH at equivalence. The pKa of hydrocyanic acid is 9.21 . Round your answer to 2 decimal places.

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Answer from: Wemaybewrong

Answer : The pH at equivalence is, 11.30

Explanation : Given,

Concentration of HCN = 0.3444 M

Volume of HCN = 100.0 mL = 0.1 L (1 L = 1000 mL)

First we have to calculate the moles of HCN

$\text{Moles of }HCN=\text{Concentration of }HCN\times \text{Volume of }HCN$

$\text{Moles of }HCN=0.3444M\times 0.1L=0.03444mol$

As we known that at equivalent point, the moles of HCN and KOH are equal.

So, Moles of KOH = Moles of HCN = 0.03444 mol

Now we have to calculate the volume of KOH.

$\text{Volume of }KOH=\frac{\text{Moles of }KOH}{\text{Concentration of }KOH}$

$\text{Volume of }KOH=\frac{0.03444mol}{0.8414M}$

$\text{Volume of }KOH=0.0409L$

Total volume of solution = 0.1 L + 0.0409 L = 0.1409 L

Now we have to calculate the concentration of KCN.

The balanced equilibrium reaction will be:

$HCN+KOH\rightleftharpoons KCN+H_2O$

Moles of KCN = 0.03444 mol

$\text{Concentration of }KCN=\frac{0.03444mol}{0.1409L}=0.244M$

At equivalent point,

$pH=\frac{1}{2}[pK_w+pK_a+\log C]$

Given:

$pK_w=14\\\\pK_a=9.21\\\\C=0.244M$

Now put all the given values in the above expression, we get:

$pH=\frac{1}{2}[14+9.21+\log (0.244)]$

pH=11.30

Therefore, the pH at equivalence is, 11.30

Answer from: Quest

Answer: abeexplanation:

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.

Answer from: Quest

"the periodic table is arranged in order of atomic number."

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A chemist titrates 100.0mL of a 0.3444M hydrocyanic acid HCN solution with 0.8414M KOH solution at 2...

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