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DanielleElmas [232]
3 years ago
6

What is the poH of a 2.6 x 10-6 M H+ solution?

Chemistry
1 answer:
andriy [413]3 years ago
4 0

Answer:

Explanation:

pH and pOH.....

The pH is a way of expressing the hydrogen ion concentration.

pH = -log[H+] ............. where [x] means "the concentration of x in moles per liter."

From pH you can compute pOH since at 25C pH + pOH = 14.00 .......... (but only at 25C)

pH = -log(2.6x10^-6) = 5.585 ..... which should be rounded to two significant digits: pH = 5.59

When taking the log of a number, only the digits to the right of the decimal reflect the precision in the original number. Since 2.6x10^-6 has two significant digits, a pH of 5.59 has two significant digits.

pOH + pH = 14.00

pOH = 14.00 - pH = 14.00 - 5.59 = 8.41 ......... at 25C

We can also use the H+ ion concentration to get the hydroxide ion concentration and from that the pOH.

Kw = [H+][OH-] = 1.00x10^-14 .......... at 25C .... like any Kc, the value changes with temperature

[OH-] = Kw / [H+] = 1.00x10^-14 / 2.6x10^-6 = 3.846x10^-9 .... to a couple of guard digits

pOH = -log[OH-] = -log(3.846x10^-9) = 8.415 ...... round to two significant digits: pOH = 8.42 ..... at 25C

=========

Just for grins, you might want to know how Kw changes with temperature, and how [H+] and [OH-] are related at some other temperatures. The pH is the pH of a neutral solution at various temperatures. For instance at 10C a neutral solution has a pH of 7.27. That's not a basic pH. 7.27 is the pH of a neutral solution, but at a different temperature. In a neutral solution at 10C [H+] = [OH-] = 5.41x10^-8M.

pH and Kw for a neutral solution at different temperatures

T .........pH ......... Kw

0......... 7.47....... 0.114 x 10-14

10....... 7.27....... 0.293 x 10-14

20....... 7.08....... 0.681 x 10-14

25....... 7.00....... 1.008 x 10-14

30....... 6.92....... 1.471 x 10-14

40....... 6.77....... 2.916 x 10-14

50....... 6.63....... 5.476 x 10-14

100..... 6.14....... 51.3 x 10-14

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The question is incomplete, the complete question is:

The solubility of slaked lime, Ca(OH)_2, in water is 0.185 g/100 ml. You will need to calculate the volume of 2.50\times 10^{-3}M HCl needed to neutralize 14.5 mL of a saturated

<u>Answer:</u> The volume of HCl required is 290mL, the mass of Ca(OH)_2 is 0.0268g, the moles of

<u>Explanation:</u>

Given values:

Solubility of Ca(OH)_2 = 0.185 g/100 mL

Volume of Ca(OH)_2 = 14.5 mL

Using unitary method:

In 100 mL, the mass of Ca(OH)_2 present is 0.185 g

So, in 14.5mL. the mass of Ca(OH)_2 present will be =\frac{0.185}{100}\times 14.5=0.0268g

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

Given mass of Ca(OH)_2 = 0.0268 g

Molar mass of Ca(OH)_2 = 74 g/mol

Plugging values in equation 1:

\text{Moles of }Ca(OH)_2=\frac{0.0268g}{74g/mol}=0.000362 mol

Moles of OH^- present = (2\times 0.000362)=0.000724mol

The chemical equation for the neutralization of calcium hydroxide and HCl follows:

Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O

By the stoichiometry of the reaction:

Moles of OH^- = Moles of H^+ = 0.000724 mol

The formula used to calculate molarity:

\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}} .....(2)

Moles of HCl = 0.000724 mol

Molarity of HCl = 2.50\times 10^{-3}

Putting values in equation 2, we get:

2.50\times 10^{-3}mol=\frac{0.000724\times 1000}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.000725\times 1000}{2.50\times 10^{-3}}=290mL

Hence, the volume of HCl required is 290mL, the mass of Ca(OH)_2 is 0.0268g, the moles of

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