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Anika [276]
3 years ago
15

Can Magnesium hydroxide be used as the base for this titration? Why or why not? 5. Describe the procedure that you will follow t

o prepare 250 ml of a .100 M soliution of KHP. 6. 25.0 ml of .100M KHP is required to neutralize 10.0 ml of barium hydroxide. Find the molarity of the barium hydroxide solution.
Chemistry
1 answer:
kvasek [131]3 years ago
4 0

Explanation:

A.

Yes

B.

2KHP + Mg(OH)2 = 2KOH + Mg(HP)2

KHP is an acidic salt which reacts with Mg(OH)2 which is a base via double displacement/decomposition reaction.

C.

• One approach is to prepare the solution volumetrically using KHP crystals.

• Preparing 1 liter of 0.1 M KHP you would add 0.1 moles of KHP

Molar mass of KHP(C8H5KO4)

= (12*8) + (5*1) + 39 + (16*4)

= 204 g/mol

Number of moles = mass/molar mass

= 20.4/204

= 0.1 mol.

• This is added to a 1 liter volumetric flask, add deionized water until near the fill line, stopper and mix

D.

Number of moles = molar concentration * volume

= 0.025 * 0.100

= 0.0025 mol

Equation for the reaction:

Ba(OH)2 + 2KHP --> Ba(KP)2 + 2H2O

1 mole of Ba(OH)2 reacted with 2 moles of KHP. By stoichiometry,

Number of moles of Ba(OH)2 = 0.0025/2

= 0.00125 mol.

Molarity = number of moles/volume

= 0.00125/0.01

= 0.125 M of Ba(OH)2.

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A 10.21 mol sample of argon gas is maintained in a 0.7564 L container at 296.9 K. What is the pressure in atm calculated using t
soldi70 [24.7K]

Answer:

The pressure in atm calculated using the van der Waals' equation, is 337.2atm

Explanation:

This is the Van der Waals equation for real gases:

(P + a/v² ) ( v-b) = R .T

where P is pressure

v is Volume/mol

R is the gas constant and T, T° in K

a y b are constant for each gas, so those values are data, from the statement.

[P + 1.345 L²atm/mol² / (0.7564L/10.21mol)² ] (0.7564L/10.21mol - 3.219×10-2 L/mol ) = 0.082 L.atm/mol.K  .  296.9K

[P + 1.345 L²atm/mol² / 5.48X10⁻³ L²/mol²] (0.074 L/mol - 3.219×10-2 L/mol ) = 0.082 L.atm/mol.K  .  296.9K

(P + 245.05 atm) (0.04181L/mol) = 0.082 L.atm/mol.K  .  296.9K

(P + 245.05 atm) (0.04181L/mol) = 24.34 L.atm/mol

0.04181L/mol .P + 10.24 L.atm/mol = 24.34 L.atm/mol

0.04181L/mol .P = 24.34 L.atm/mol - 10.24 L.atm/mol

0.04181L/mol. P = 14.1 L.atm/mol

P = 14.1 L.atm/mol / 0.04181 mol/L

P = 337.2 atm

4 0
3 years ago
The ka for hcn is 4.9 ⋅ 10-10. What is the value of kb for cn-?
marusya05 [52]

Answer:

kb = 2,0x10⁻⁵

Explanation:

The ka for HCN is:

HCN ⇄ H⁺ + CN⁻; ka = 4,9x10⁻¹⁰ <em>(1)</em>

The inverse reaction has an equilibrium constant of:

H⁺ + CN⁻ ⇄ HCN k = 1/4,9x10⁻¹⁰ = 2,0x10⁹ <em>(2)</em>

As the equilibrium of the water is:

H₂O ⇄ H⁺ + OH⁻; kw = 1x10⁻¹⁴ <em>(3)</em>

The sum of (2) + (3) gives:

H₂O + CN⁻ ⇄ HCN + OH⁻; kb = kw×k = 1x10⁻¹⁴×2,0x10⁹ =

2,0x10⁻⁶; <em>kb = 2,0x10⁻⁵</em>

<em />

<em>-In fact, the general formula to convert from ka to kb is:</em>

<em>kb = kw / ka-</em>

<em />

I hope it helps!

5 0
3 years ago
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