what are the hydrogen ion and hydroxide ion concentrations in water? what is the pH and pOH of pure water?

Chemistry

1 answer:

Bumek [7]2 years ago

3 0

Answer:

The pH of pure water is 7, the negative logarithm of 1 X 10-7. A neutral solution is one that is neither acidic nor basic. The hydrogen ion concentration equals the hydroxide ion concentration, and both equal 1 X 10-7 M. In a neutral solution, then, pH = pOH = 7.

Explanation:

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A chemist fills a reaction vessel with 3.82 atm methanol (CH,OH) gas, 7.56 am oxygen (O2) gas, 5.29 atm carbon dioxide (CO2) gas

alisha [4.7K]

Answer: The Gibbs free energy of the given reaction is $1.379\times 10^3kJ$

Explanation:

The equation used to calculate Gibbs free energy change is of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

For the given chemical reaction:

$2CH_3OH(g)+3O_2(g)\rightarrow 2CO_2(g)+4H_2O(g)$

The equation for the Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(CO_2(g))})+(4\times \Delta G^o_f_{(H_2O(g))})]-[(2\times \Delta G^o_f_{(CH_3OH(g))})+(3\times \Delta G^o_f_{(O_2(g))})]$

We are given:

$\Delta G^o_f_{(H_2O(g))}=-228.57kJ/mol\\\Delta G^o_f_{(CO_2(g))}=-394.36kJ/mol\\\Delta G^o_f_{(CH_3OH(g))}=-161.96kJ/mol\\\Delta G^o_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-394.36))+(4\times (-228.57))]-[(2\times (-161.96))+(3\times (0))]\\\\\Delta G^o_{rxn}=-1379.08kJ/mol$

The equation used to Gibbs free energy of the reaction follows:

$\Delta G=\Delta G^o+RT\ln Q_{p}$

where,

$\Delta G$ = free energy of the reaction

$\Delta G^o$ = standard Gibbs free energy = -1379.08 kJ/mol = -1379080 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/K mol

T = Temperature = $25^oC=[273+25]K=298K$

$Q_{p}$ = Ratio of concentration of products and reactants = $\frac{(p_{CO_2})^2(p_{H_2O})^4}{(p_{CH_3OH})^2(p_{O_2})^3}$

$p_{CO_2}=5.29atm\\p_{H_2O}=3.89atm\\p_{CH_3OH}=3.82atm\\p_{O_2}=7.56atm$

Putting values in above expression, we get:

$\Delta G=-1379080J/mol+(8.314J/K.mol\times 298K\times \ln (\frac{(5.29)^2\times (3.89)^4}{(3.82)^2\times (7.56)^3}))\\\\\Delta G=-1379039J=1379.039kJ=1.379\times 10^3kJ$