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fgiga [73]
3 years ago
14

Calculate the radius ratio for NaBr if the ionic radii of Na + and Br − are 102 pm and 196 pm , respectively. radius ratio: Base

d on the radius ratio, what is the expected coordination number of NaBr ? The range of radius ratios for each coordination number is given in parentheses.

Chemistry
1 answer:
Fudgin [204]3 years ago
8 0

Answer : The expected coordination number of NaBr is, 6.

Explanation :

Cation-anion radius ratio : It is defined as the ratio of the ionic radius of the cation to the ionic radius of the anion in a cation-anion compound.

This is represented by,

\frac{r_{cation}}{r_{anion}}

When the radius ratio is greater than 0.155, then the compound will be stable.

Now we have to determine the radius ration for NaBr.

Given:

Radius of cation, Na^+ = 102 pm

Radius of cation, Br^- = 196 pm

\frac{r_{cation}}{r_{anion}}=\frac{102}{196}=0.520

As per question, the radius of cation-anion ratio is between 0.414-0.732. So, the coordination number of NaBr will be, 6.

The relation between radius ratio and coordination number are shown below.

Therefore, the expected coordination number of NaBr is, 6.

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Calculate the change in pH when 71.0 mL of a 0.760 M solution of NaOH is added to 1.00 L of a solution that is 1.0O M in sodium
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Explanation:

It is known that pK_{a} value of acetic acid is 4.74. And, relation between pH and pK_{a} is as follows.

                    pH = pK_{a} + log \frac{[CH_{3}COOH]}{[CH_{3}COONa]}

                          = 4.74 + log \frac{1.00}{1.00}

So, number of moles of NaOH = Volume × Molarity

                                                   = 71.0 ml × 0.760 M

                                                    = 0.05396 mol

Also, moles of  CH_{3}COOH = moles of CH_{3}COONa

                                          = Molarity × Volume

                                          = 1.00 M × 1.00 L

                                          = 1.00 mol

Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.

            CH_{3}COONa + NaOH \rightarrow CH_{3}COOH

Initial :    1.00 mol                                  1.00 mol

NaoH addition:               0.05396 mol

Equilibrium : (1 - 0.05396 mol)    0           (1.00 + 0.05396 mol)

                    = 0.94604 mol                       = 1.05396 mol

As, pH = pK_{a} + log \frac{[CH_{3}COONa]}{[CH_{3}COOH]}

               = 4.74 +  log \frac{0.94604}{1.05396}

               = 4.69

Therefore, change in pH will be calculated as follows.

                         pH = 4.74 - 4.69

                               = 0.05

Thus, we can conclude that change in pH of the given solution is 0.05.

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