3 years ago

The element in period 4 and group 8

Chemistry

1 answer:

Leno4ka [110]3 years ago

3 0

I am pretty sure that the element is Iron (Fe)

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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

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

Moles of base

$0.01202\times100\times\dfrac{1}{10^3}=0.001202\ \text{moles}$

Mass of base is given by

$0.001202\times 322.2=0.387\ \text{g}$

The required mass of the base is 0.387 g.

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