Chemistry, 09.12.2019 18:31 skyrileycasting

What is the ph of a sodium formate solution prepared by adding 0.680 grams of sodium formate to 100.0 ml of water at 25.0°c? the ka at 25.0 °c for formic acid is 1.8 x 10-4. enter your answer with two decimal places.

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Answer from: sofipalis9869

Answer : The pH of the solution is, 8.38

Solution : Given,

Concentration (c) = 0.11 M

Acid dissociation constant = $k_a=1.8\times 10^{-4}$

First we have to calculate the moles of sodium formate.

$Moles=\frac{Mass}{\text{Molar mass}}=\frac{0.680g}{68g/mole}=0.010mol$

Now we have to calculate the concentration of formic acid.

Concentration of formic acid = Concentration of sodium formate = $\frac{Moles}{Volume}=\frac{0.010mol}{0.100L}=0.1M$

Now we have to calculate the base dissociation constant.

$k_w=k_a\times k_b\\\\k_b=\frac{k_w}{k_a}\\\\k_b=\frac{1\times 10^{-14}}{1.8\times 10^{-4}}\\\\k_b=5.6\times 10^{-11}$

The equilibrium reaction for dissociation of (weak base) is,

$HCOO^-+H_2O\rightleftharpoons HCOOH+OH^-$

initially conc. 0.1 0 0

At eqm. (0.1-x) x x

First we have to calculate the value of 'x'.

Formula used :

$k_b=\frac{[HCOOH][OH^-]}{[HCOO^-]}$

Now put all the given values in this formula ,we get:

$5.6\times 10^{-11}=\frac{(x)(x)}{(0.1-x)}$

By solving the terms, we get

$x=2.4\times 10^{-6}M$

Thus, the concentration of hydroxide ion is:

$[OH^-]=x=2.4\times 10^{-6}M$

Now we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (2.4\times 10^{-6})$

pOH=5.62

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-5.62\\\\pH=8.38$

Therefore, the pH of the solution is, 8.38

Answer from: Quest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.