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larisa [96]
3 years ago
11

For the following reaction, label each of the below species as an acid or a base. Use lower case letters only (e.g. acid)

Chemistry
1 answer:
s2008m [1.1K]3 years ago
5 0

Explanation:

According to Arrhenius, bases are the species which dissociate to give hydroxide ions, that is, OH^{-}.  

For example, NaOH + H_{2}O \rightarrow Na^{+} + OH^{-}

Arrhenius acids are the species which dissociate to give hydrogen ions, that is, H^{+}.

For example, CH_{3}COOH + H_{2}O \rightleftharpoons H_{3}O^{+} + CH_{3}COO^{-}

In the given example, HCN is donating the hydrogen ion therefore, it is an acid whereas HPO^{2-}_{4} is accepting the hydrogen ions so it acts as a base.

Now, standard \Delta G^{o} values of the given species are as follows.

 HCN = 124.7 kJ/mol,      HPO^{2-}_{4} = -1089.3 kJ/mol

 H_{2}PO^{-}_{4} = -1130.4 kJ/mol,       CN^{-} = 172.4 kJ/mol

Therefore, \Delta G^{o} of the given reaction will be calculated as follows.

      \Delta G^{o} = \Delta G_{products} - \Delta G_{reactants}

                   = [-1130.4 + 172.4] - [124.7 + (-1089.3)]

                   = (-958 + 964.6) kJ/mol

                   = 6.6 kJ/mol

Since, \Delta G^{o} is greater than zero. Hence, the reaction will be non-spontaneous.

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g In animal tissues the rate of conversion of pyruvate to acetyl-CoA is regulated by the ratio of phosphorylated and dephosphory
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Which ion has the same electron configuration as an atom of He?
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-Helium atom has two electrons, therefore, it is similar with an negatively charged hydrogen.

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2 years ago
A chemist titrates 220.0 mL of a 0.1917M propionic acid (HC2H5CO2) solution with 0.1787 M KOH solution at 25°C. Calculate the p
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Answer : The pH at equivalence is, 9.08

Explanation : Given,

Concentration of HC_2H_5CO_2 = 0.1917 M

Volume of HC_2H_5CO_2 = 220.0 mL = 0.220 L (1 L = 1000 mL)

First we have to calculate the moles of HC_2H_5CO_2

\text{Moles of }HC_2H_5CO_2=\text{Concentration of }HC_2H_5CO_2\times \text{Volume of }HC_2H_5CO_2

\text{Moles of }HC_2H_5CO_2=0.1917M\times 0.220L=0.0422

As we known that at equivalent point, the moles of HC_2H_5CO_2 and KOH are equal.

So, Moles of KOH = Moles of HC_2H_5CO_2 = 0.0422 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{0.0422mol}{0.1787M}

\text{Volume of }KOH=0.00754

Total volume of solution = 0.220 L + 0.00754 L = 0.22754 L

Now we have to calculate the concentration of KCN.

The balanced equilibrium reaction will be:

HC_2H_5CO_2+KOH\rightleftharpoons C_2H_5CO_2K+H_2O

Moles of C_2H_5CO_2K = 0.0422 mol

\text{Concentration of }C_2H_5CO_2K=\frac{0.0422mol}{0.22754L}=0.1855M

At equivalent point,

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

Given:

pK_w=14\\\\pK_a=4.89\\\\C=0.1855M

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

pH=\frac{1}{2}[14+4.89+\log (0.1855)]

pH=9.08

Therefore, the pH at equivalence is, 9.08

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