The amount of W(OH)2 needed would be 448.126 g
<h3>Stoichiometric calculation</h3>
From the equation of the reaction:
W(OH)2 + 2 HCl → WCl2 + 2 H2O
The mole ratio of W(OH)2 to HCl is 1:2
Mole of 150g HCl = 150/36.461
= 4.11 moles
Equivalent mole of W(OH)2 = 4.11/2
= 2.06 moles
Mass of 2.06 moles W(OH)2 = 2.06 x 217.855
= 448.188g
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Answer : The maximum amount of nickel(II) cyanide is 
Explanation :
The solubility equilibrium reaction will be:
Initial conc. 0.220 0
At eqm. (0.220+s) 2s
The expression for solubility constant for this reaction will be,
![K_{sp}=[Ni^{2+}][CN^-]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BNi%5E%7B2%2B%7D%5D%5BCN%5E-%5D%5E2)
Now put all the given values in this expression, we get:
Therefore, the maximum amount of nickel(II) cyanide is
Answer:
C. CH3COOH, Ka = 1.8 E-5
Explanation:
analyzing the pKa of the given acids:
∴ pKa = - Log Ka
A. pKa = - Log (1.0 E-3 ) = 3
B. pKa = - Log (2.9 E-4) = 3.54
C. pKa = - Log (1.8 E-5) = 4.745
D. pKa = - Log (4.0 E-6) = 5.397
E. pKa = - Log (2.3 E-9) = 8.638
We choose the (C) acid since its pKa close to the expected pH.
⇒ For a buffer solution formed from an acid and its respective salt, we have the equation Henderson-Hausselbach (H-H):
- pH = pKa + Log ([CH3COO-]/[CH3COOH])
∴ pH = 4.5
∴ pKa = 4.745
⇒ 4.5 = 4.745 + Log ([CH3COO-]/[CH3COOH])
⇒ - 0.245 = Log ([CH3COO-]/[CH3COOH])
⇒ 0.5692 = [CH3COO-]/[CH3COOH]
∴ Ka = 1.8 E-5 = ([H3O+].[CH3COO-])/[CH3COOH]
⇒ 1.8 E-5 = [H3O+](0.5692)
⇒ [H3O+] = 3.1623 E-5 M
⇒ pH = - Log ( 3.1623 E-5 ) = 4.5
