Answer:
[O2(g)][SO2(g)]^2/[SO3(g)]^2
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>
2 hydrogen molecules to one oxygen molecule
Remember by H2O since O doesn’t have a number after it you know there is only one oxygen molecule
<span>using the law pv=nrT and equating these you get the equation v1/t1 = v2/t2 since pressure is constant it also cancels with n and r. show that v1=36.4, t1 = 25 + 273.15 and t2 = 88 +273.15. 273.15 is the Kelvin conversion. then solve for v2. This is 44.1 L.</span>
