Answer:
Molarity of the sodium hydroxide solution is 1.443 M/L
Explanation:
Given;
0.60 M concentration of NaOH contains 2.0 L
3.0 M concentration of NaOH contains 495 mL
Molarity is given as concentration of the solute per liters of the solvent.
If the volumes of the two solutions are additive, then;
the total volume of NaOH = 2 L + 0.495 L = 2.495 L
the total concentration of NaOH = 0.6 M + 3.0 M = 3.6 M
Molarity of NaOH solution = 3.6 / 2.495
Molarity of NaOH solution = 1.443 M/L
Therefore, molarity of the sodium hydroxide solution is 1.443 M/L
Explanation:When the stove turns on the skillet will become very hot because of the conduction of heat transferring from the stove to the skillet.
Answer:
pH = 2.69
Explanation:
The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>
<em />
The reaction of HNO₂ with KOH is:
HNO₂ + KOH → NO₂⁻ + H₂O + K⁺
Moles of HNO₂ and KOH that react are:
HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>
KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>
That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:
NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻
HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂
It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:
pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]
pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]
<h3>pH = 2.69</h3>
Answer:
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