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:
C3H8 +5O2 arrow 3CO2 +4H2O
Answer:
0.007 mol
Explanation:
We can solve this problem using the ideal gas law:
PV = nRT
where P is the total pressure, V is the volume, R the gas constant, T is the temperature and n is the number of moles we are seeking.
Keep in mind that when we collect a gas over water we have to correct for the vapor pressure of water at the temperature in the experiment.
Ptotal = PH₂O + PO₂ ⇒ PO₂ = Ptotal - PH₂O
Since R constant has unit of Latm/Kmol we have to convert to the proper unit the volume and temperature.
P H₂O = 23.8 mmHg x 1 atm/760 mmHg = 0.031 atm
V = 1750 mL x 1 L/ 1000 mL = 0.175 L
T = (25 + 273) K = 298 K
PO₂ = 1 atm - 0.031 atm = 0.969 atm
n = PV/RT = 0.969 atm x 0.1750 L / (0.08205 Latm/Kmol x 298 K)
n = 0.007 mol
I believe it is C, but I can't guarantee it.
