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rodikova [14]
2 years ago
5

Consider a buffer solution containing CH3COOH and CH3COO-, with an equilibrium represented by: CH3COOH(aq) + H2O (l) ←----→ H3O+

(aq) + CH3COO- (aq) Describe what occurs if a strong acid such as HNO3 is added to the system, including an explanation of the direction of the equilibrium shift. Describe what occurs if a strong base such as KOH is added, including an explanation of the direction of the equilibrium shift.
Chemistry
1 answer:
Irina-Kira [14]2 years ago
5 0

Answer:

Here's what I get.

Explanation:

(a) The buffer equilibrium

The equation for the buffer equilibrium is

\rm CH_{3}COOH(aq) + H$_{2}$O(l) $\, \rightleftharpoons \,$ CH$_{3}$COO$^{-}$(aq) + H$_{3}$O$^{+}$(aq)

(b) Addition of acid

If you add a strong acid like HNO₃, you are increasing the concentration of hydronium ion.

Per Le Châtelier's Principle, the system will respond in such a way as to decrease the concentration of hydronium ion.

The position of equilibrium will shift to the left.

(c) Addition of base.

If you add a strong base like KOH, The hydroxide ions will react with the hydronium ions to form water.

The concentration of hydronium ions will decrease.

Per Le Châtelier's Principle, the system will respond in such a way as to increase the concentration of hydronium ions.

The position of equilibrium will shift to the right.

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During a titration the following data were collected. A 10. mL portion of an unknown monoprotic acid solution was titrated with
Liula [17]

Answer:

8.0 moles

Explanation:

Since the acid is monoprotic, 1 mole of the acid will be required to stochiometrically react with 1 mole of NaOH.

Using the formula: \frac{concentration of acid X volume of acid}{concentration of base X volume of base} = \frac{mole of acid}{mole of base}

Concentration of acid = ?

Volume of acid = 10 mL

Concentration of base = 1.0 M

Volume of base = 40 mL

mole of acid = 1

mole of base = 1

Substitute into the equation:

\frac{concentration of acid X 10}{1.0 X 40} = \frac{1}{1}

Concentration of acid = 40/10 = 4.0 M

To determine the number of moles of acid present in 2.0 liters of the unknown solution:

Number of moles = Molarity x volume

molarity = 4.0 M

Volume = 2.0 Liters

Hence,

Number of moles = 4.0 x 2.0 = 8 moles

8 0
2 years ago
For a ternary solution at constant T and P, the composition dependence of molar property M is given by: M = x1M1 + x2M2 + x3M3 +
AveGali [126]

Answer:

M_{i} = M_{i} + C_{xjxk} (1-2x_{i}) ...1

M^{\alpha } = M_{i} + CX_{xjxk}          ...2

Explanation:

The ternary constant is given by the following equation:

The symbol XiXi, where XX is an extensive property of a homogeneous mixture and the subscript ii identifies a constituent species of the mixture, denotes the partial molar quantity of species ii defined by

M_{i}  = [\frac{d(nM)}{dn_{i} }]_{P,t,n,j}

This is the rate at which property  X  changes with the amount of species  i  added to the mixture as the temperature, the pressure, and the amounts of all other species are kept constant.  A partial molar quantity is an intensive state function.  Its value depends on the temperature, pressure, and composition of the mixture.

In a multi phase system (in this case, a ternary system), the components resolved give:

M_{i} = M_{i} + C_{xjxk} (1-2x_{i})

and M^{\alpha } = M_{i} + CX_{xjxk}

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2 years ago
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creativ13 [48]

Answer:

O FX will be greater than FY

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This required force is proportional to the liquid's surface tension. This means that the higher the surface tension, the higher the required force to stretch it is.

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5.00 g of hydrogen gas and 50.0g of oxygen gas are introduced into an otherwise empty 9.00L steel cylinder, and the hydrogen is
GenaCL600 [577]
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2H2 + O2 -> 2H20

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2) Reactant quantities converted to moles

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Total number of moles: 0.3125mol + 2.5 mol = 2.8125 mol

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Use pV = nRT
n = 2.8125
V= 9 liters
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p = nRT/V  = 7.9 atm
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