What volume of 0.25 M HCl would be needed to neutralize 45 mL of 0.12 M solution of Mg(OH)₂
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Answer:
0.4 M
Explanation:
The process that takes place in an aqueous K₂HPO₄ solution is:
- K₂HPO₄ → 2K⁺ + HPO₄⁻²
First we <u>calculate how many K₂HPO₄ moles are there in 200 mL of a 0.2 M solution</u>:
- 200 mL * 0.2 M = 40 mmol K₂HPO₄
Then we <u>convert K₂HPO₄ moles into K⁺ moles</u>, using the <em>stoichiometric coefficients</em> of the reaction above:
- 40 mmol K₂HPO₄ *
= 80 mmol K⁺
Finally we <em>divide the number of K⁺ moles by the volume</em>, to <u>calculate the molarity</u>:
- 80 mmol K⁺ / 200 mL = 0.4 M