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1 year ago

determine whether the aqueous solutions of each of the following salts will be acidic, basic, or neutral

Chemistry

1 answer:

eduard1 year ago

6 0

KCl is neutral because KOH is a strong base and HCl is a strong acid

To determine if the salt is neutral, acidic or basic, determine how the salt was made, or from which acid and base it was derived. Once we have determined how the salt was made, determine if the acid and base are both strong, weak or one of each. If both are strong (e.g. NaOH + HCl), then the salt (NaCl) is NEUTRAL. If the base is strong and the acid is weak (KOH + HCN), then the salt (KCN) is BASIC. IF the base is weak and the acid is strong (NH3 + HCl), then the salt (NH4Cl) is ACIDIC. If both the base and the acid are weak (NH3 + CH3COOH), then then salt (CH3COONH4) will have a pH determined by which is stronger (the acid or the base) and this will depend on the Ka of the weak acid compared to the Kb of the weak base.

KCl is neutral because KOH is a strong base and HCl is a strong acid NaCN is generally basic because NaOH is a strong base and HCN is a very week acid. NH4NO3 is usually acid because HNO3 is a strong acid and NH4OH is a week base

CH3COONH4 is exactly neutral because both NH4OH ad CH3COOH are respectively week base and week acid. Moreover, the Ka of acetic acid is exactly equal to the Kb of NH4OH

To know about strong base.

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The enthalpy of solution (∆H) of KOH is -57.6 kJ/mol. If 3.66 g KOH is dissolved in enough water to make a 150.0 mL solution, wh

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When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

$3.66g \times \frac{1mol}{56.11g} \times \frac{(-57.6kJ)}{mol} = -3.76 kJ$

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

$Qr+Qa = 0\\\\Qa = -Qr = 3.76 kJ$

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

$150.0 mL \times \frac{1.02g}{mL} = 153 g$

Given the specific heat capacity of the solution (c) is 4.184 J/g･°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

$Qa = c \times m \times \Delta T\\\\\Delta T = \frac{Qa}{c \times m} = \frac{3.76 \times 10^{3}J }{\frac{4.184J}{g.\° C } \times 153g} = 5.87 \° C$

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

Learn more: brainly.com/question/4400908

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