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rodikova [14]
3 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]3 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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When 91.5 g of isopropyl alcohol which has an empirical formula of C3H8O is burned in excess oxygen gas, how many grams of H2O a
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Answer:

109.7178g of H2O

Explanation:

First let us generate a balanced equation for the reaction. This is illustrated below:

2C3H8O + 9O2 —> 6CO2 + 8H2O

Next we will calculate the molar mass and masses of C3H8O and H20. This is illustrated below:

Molar Mass of C3H8O = (3x12.011) + (8x1.00794) + 15.9994 = 36.033 + 8.06352 + 15.9994 = 60.09592g/mol.

Mass of C3H8O from the balanced equation = 2 x 60.09592 = 120.19184g

Molar Mass of H2O = (2x1.00794) + 15.9994 = 2.01588 + 15.9994 = 18.01528g/mol

Mass of H2O from the balanced equation = 8 x 18.01528 = 144.12224g

From the equation,

120.19184g of C3H8O produced 144.12224g of H20.

Therefore, 91.5g of C3H8O will produce = (91.5 x 144.12224) /120.19184 = 109.7178g of H2O

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