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1 year ago

What is the ph of an aqueous solution with a hydrogen ion concentration of [H+]=1. 2×10^−8 m?

Chemistry

1 answer:

il63 [147K]1 year ago

4 0

The pH of an aqueous solution with a hydrogen ion concentration of [H+]=1. 2×10^−8 m is 7.92.

The entire form of pH is potential of Hydrogen. pH is called the poor logarithm of H+ ion attention. hence the that means of the name pH is explained because the energy of hydrogen. pH describes the concentration of the hydrogen ions in a solution and it's miles the indicator of acidity or basicity.

pH is an important amount that reflects the chemical conditions of an answer. The pH can manage the provision of vitamins, organic features, microbial activity, and the conduct of chemical substances.

pH stands for capacity hydrogen and it's used to explain the acidity or alkalinity of a substance. The pH scale is measured from 1 to fourteen. 7 is impartial and the lower the quantity, the extra acidic and the higher, the greater alkaline.

explaination,

[H+]=1. 2×10^−8 m

log[h+] = 1. 2×10^−8 m

= 8 - log (1.2)

= 8 - 0.079

PH = 7.921

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Read 2 more answers

The final volume of buffer solution must be 100.00 mL and the final concentration of the weak acid must be 0.100 M. Based on thi

Grace [21]

Answer:

0.387 g

Explanation:

pH of the buffer = 1

V = Volume of solution = 100 mL

[HA] = Molarity of HA = 0.1 M

$K_a$ = Acid dissociation constant = $1.2\times 10^{-2}$

(assuming base as $Na_2SO_410H_2O$ )

Molar mass of base = 322.2 g/mol

pKa is given by

$pK_a=-\log K_a\\\Rightarrow pKa=-\log(1.2\times 10^{-2})\\\Rightarrow pK_a=1.92$

From the Henderson-Hasselbalch equation we get

$pH=pK_a+\log\dfrac{[A^-]}{[HA]}\\\Rightarrow pH-pK_a=\log\dfrac{[A^-]}{[HA]}\\\Rightarrow 10^{pH-pK_a}=\dfrac{[A^-]}{[HA]}\\\Rightarrow [A^-]=10^{pH-pK_a}[HA]\\\Rightarrow [A^-]=10^{1-1.92}\times0.1\\\Rightarrow [A^-]=0.01202\ \text{M}$