Question

Find the ph of the solution obtained by mixing 2ml hcl with ph=2 and 3ml koh with ph =12

Answer 1

Explanation:

refer to the attachment.

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Answer 2

The pH of the solution obtained by mixing 2ml hcl with pH=2 and 3ml KOH with pH =12 is 11.6

Explanation:

pH is the measure of acidity or alkalinity of a solution.

Acids have pH ranging from 1 to 6.9 and bases have pH ranging from 7.1 to 14.Neutral substances have pH of 7.

To calculate the $[H^+]$ for HCl

$pH=-log[H^+]$

$2=-log[H^+]$

$[H^+]=10^{-2}M$

Moles of $H^+$ ion = $Molarity\times {\text {Volume in L}}=10^{-2}M\times 0.002ml=2\times 10^{-5}mol$

To calculate the $[OH^-]$ for KOH

pH+pOH = 14

pOH=14-12= 2

$pOH=-log[OH^-]$

$2=-log[OH^-]$

$[OH^-]=10^{-2}M$

Moles of $OH^-$ ion = $Molarity\times {\text {Volume in ml}}=10^{-2}M\times 0.003ml=3\times 10^{-5}mol$

The chemical equation for the reaction of NaOH with sulfuric acid follows:

$HCl+KOH\rightarrow KCl+H_2O$

For neutralization:

1 mole of $H^+$ ion will react with 1 mole of $OH^-$ ion

$2\times 10^{-5}mol$ mole of $H^+$ ion will react with $2\times 10^{-5}mol$ mole of $OH^-$ ion

Thus $3\times 10^{-5}-2\times 10^{-5}$ = $1\times 10^{-5}mol$ moles of KOH are left in 0.005 L of solution

$[OH^-]=\frac{2\times 10^{-5}mol}{0.005L}=4\times 10^{-3}M$

$pOH=-log[4\times 10^{-3}]=2.4$

$pH=14-2.4=11.6$

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