An acid with a H+ molarity of 1x10^-5 M has a ph of?

Chemistry

1 answer:

Pavlova-9 [17]3 years ago

8 0

Answer:

The answer to your question is pH = 5

Explanation:

Data

[H⁺] = 1 x 10⁻⁵

pH = ?

pH is a measure of the hydrogen concentration. It measures the acidity or alkalinity. The range of pH goes from 0 to 14. The pH for an acid is from 0 to 6.9, for a neutral solution is 7.0 and for an alkali is from 7.1 to 14.

Formula

pH = -log [H⁺]

-Substitution

pH = -log [1 x 10⁻⁵]

-Simplification

pH = - (-5)

-Result

pH = 5

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Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang

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Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

$\Delta (H_{f})_{MnO_{2}}$ = -520.0 KJ/mole

$\Delta (H_{f})_{Al_{2}O_{3}}$ = -1699.8 KJ/mole

The balanced chemical reaction is,

$3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)$

Formula used :

$\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}$

$\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))$

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

$\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]$