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3 years ago

How do you know the ph level of 10^-8

Chemistry

1 answer:

sergiy2304 [10]3 years ago

5 0

pH=6.98

Explanation:

This is a very interesting question because it tests your understanding of what it means to have a dynamic equilibrium going on in solution.

As you know, pure water undergoes self-ionization to form hydronium ions, H3O+, and hydroxide anions, OH−.

2H2O(l]⇌H3O+(aq]+OH−(aq]→ very important!

At room temperature, the value of water's ionization constant, KW, is equal to 10−14. This means that you have

KW=[H3O+]⋅[OH−]=10−14

Since the concentrations of hydronium and hydroxide ions are equal for pure water, you will have

[H3O+]=√10−14=10−7M

The pH of pure water will thus be

pH=−log([H3O+])

pH=−log(10−7)=7

Now, let's assume that you're working with a 1.0-L solution of pure water and you add some 10

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Explain why the concentrations of a mixture at equilibrium are constant as a function of time

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3 years ago

Given the following: [G3P] = 1.5x10-5M; [BPG] = 3.0x10-3M ; [NAD+] = 1.2x10-5M; [NADH]=1.0x10-4 ; [HPO42-]= 1.2x10-5 M; pH = 7.5

mel-nik [20]

Answer: The given reaction is non-spontaneous in nature.

Explanation:

To calculate the $H^+$ concentration, we use the equation:

$pH=-\log[H^+]$

We are given:

pH of the solution = 7.5

$7.5=-\log [H^+]$

$[H^+]=10^{-7.5)=3.1\times 10^{-8}M$

For the given chemical equation:

$\text{Glyceraldehyde3-phosphate }+NAD^++HPO_4^{2-}\rightarrow \text{1,3-Biphosphoglycerate }+NADH+H^+$

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

$\Delta G=\Delta G^o+RT\ln K_{eq}$

where,

$\Delta G$ = free energy of the reaction

$\Delta G^o$ = standard Gibbs free energy = 6.3 kJ/mol = 6300 J/mol (Conversion factor: 1kJ = 1000J)

R = Gas constant = 8.314J/K mol

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

$K_{eq}$ = Ratio of concentration of products and reactants = $\frac{[BPG][NaDH][H^+]}{[G_3P][NAD^+][HPO_4^{2-}]}$

$[BPG]=3.0\times 10^{-3}M$

$[NADH]=1.0\times 10^{-4}M$

$[H^+]=3.1\times 10^{-8}M$

$[G_3P]=1.5\times 10^{-5}M$

$[NAD^+]=1.2\times 10^{-5}M$

$[HPO_4^{2-}]=1.2\times 10^{-5}M$

Putting values in above equation, we get:

$\Delta G=6300J/mol+(8.314J/K.mol\times 298K\times \ln (\frac{(3.0\times 10^{-3})\times (1.0\times 10^{-4})\times (3.1\times 10^{-8})}{(1.5\times 10^{-5})\times (1.2\times 10^{-5})\times (1.2\times 10^{-5})}))\\\\\Delta G=9917.02J/mol=9.92kJ/mol$

For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.

As, the Gibbs free energy of the reaction is positive. The reaction is said to be non-spontaneous.

Hence, the given reaction is non-spontaneous in nature.

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3 years ago