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>
Ph= - log [H+] = -log 1.00× 10-7 = -(log 1 + log 10-7) = -( 0 + (-7log 10) = -( -7) = 7
Answer:
204.5505 grams
2.5666 moles
Explanation:
For the first question, multiply 3.5 (# of moles) by 58.443 (g/mol for NaCl).
58.443 * 3.5
<em>I'll distribute 3.5 into 58.443.</em>
(3.5 * 50) + (3.5 * 8) + (3.5 * 0.4) + (3.5 * 0.04) + (3.5 * 0.003)
175 + 28 + 1.4 + 0.14 + 0.0105
203 + 1.4 + 0.14 + 0.0105
204.4 + 0.14 + 0.0105
204.54 + 0.0105
204.5505 grams
There are 204.5505 grams in 3.5 moles of NaCl.
For the second question, divide 150 (# of grams) by 58.443 (g/mol for NaCl). I'll convert both into fractions.
150/1 * 1000/58443
150000/58443
2.56660336 moles
2.5666 moles (rounded to 4 places to keep consistency with the first answer) are in 150 grams of NaCl.
Answer:
Option b.
The triple point of a substance is the temperature and pressure conditions at which states of a substance coexist at equilibrium