P2O5 = Phosphorus pentoxide
CuO = Copper (II) oxide
NH4CI = Ammonium Chloride
Mn(OH)2 = Pyrochroite
H2O2 = Hydrogen peroxide
P4S9 = Tetraphosphorus nonasulfide
CIO2 = Chlorine dioxide
NaF = Sodium fluoride
FeSO3 = Iron (II) Sulfite
Fe(NO3)3 = Iron (III) Nitrate
Cr(NO2)3 = Chromium (III) Nitrite
NaHCO3 = Sodium Hydrogen Carbonate
H2PO4 = Dihydrogen Phosphate Ion
NaCN = Sodium Cyanide
IF7 = Iodine Heptafluoride
PCI3 = Phosphorus Trichloride
Answer:
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Explanation:
im sorry i connot answer your question
Answer:
The answer to your question is below
Explanation:
a) HCl 0.01 M
pH = -log [0.01]
pH = - (-2)
pH = 2
b) HCl = 0.001 M
pH = -log[0.001]
pH = -(-3)
pH = 3
c) HCl = 0.00001 M
pH = -log[0.00001]
pH = - (-5)
pH = 5
d) Distilled water
pH = 7.0
e) NaOH = 0.00001 M
pOH = -log [0.00001]
pOH = -(-5)
pH = 14 - 5
pH = 9
f) NaOH = 0.001 M
pOH =- log [0.001]
pOH = 3
pH = 14 - 3
pH = 11
g) NaOH = 0.1 M
pOH = -log[0.1]
pOH = 1
pH = 14 - 1
pH = 13
Answer:
[HF]₀ = 0.125M
Explanation:
NaOH + HF => NaF + H₂O
Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3. This is 0.089M NaF and 0.001M HF remaining.
=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.
HF ⇄ H⁺ + F⁻
C(eq) [HF] 10⁻³M 0.089M (<= soln after adding 20ml 0.200M NaOH)
Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka
[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M
