Answer:
11.31g NaClO₂
Explanation:
<em> Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45? </em>
It is possible to answer this question using Henderson-Hasselbalch equation:
pH = pKa + log₁₀ [A⁻] / [HA]
<em>Where pKa is -log Ka = 1.9547; [A⁻] is the concentration of the conjugate base (NaClO₂), [HA] the concentration of the weak acid</em>
You can change the concentration of the substance if you write the moles of the substances:
[Moles HClO₂] = 250mL = 0.25L×(1.60mol /L) = <em>0.40 moles HClO₂</em>
Replacing in H-H expression, as the pH you want is 1.45:
1.45 = 1.9547 + log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
-0.5047 = log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
<em>0.3128 = </em>[Moles NaClO₂] / [0.40 moles HClO₂]
0.1251 = Moles NaClO₂
As molar mass of NaClO₂ is 90.44g/mol, mass of 0.1251 moles of NaClO₂ is:
0.1251 moles NaClO₂ ₓ (90.44g / mol) =
<h3>11.31g NaClO₂</h3>
Answer:
oof idk I would suggest looking at an example and try going off that
B) Equal to the number of protons
Metals on the left side, metalloids on the staircase, nonmetals on right side
T = 14400 s
26.5 x 14400=381600 C
381600/96500=3.95 Faradays
Cu2+ + 2e- = Cu
3.95 faradays ( 1 mol/ 2 Faradays) = 1.97
mass = 1.97 x 63.55 g/mol=125 g
moles Au = 33.1 / 196.967 g/mol=0.168
Au+ + 1e- = Au
0.168 ( 1 Faraday/ 1mol)= 0.168 Faraday
0.168 x 96500=16217 Coulombs
16217 / 5.00=3243 s => 54 min