A chemist dissolves 320.mg of pure nitric acid in enough water to make up 290.mL of solution. Calculate the pH of the solution. Be sure your answer has the correct number of significant digits.

1.7

Explanation:

The formula to be used here is;

pH = -log[H⁺]

To determine the concentration of H⁺ of the nitric acid (HNO₃), formula for molarity is used

Molarity = number of  moles/volume (in dm³ or L)

number of moles (n) = given mass/molar mass

Molar mass of HNO₃ = 1 + 14 + (16 x 3)

= 63g

given mass (in grams) = 320mg⇔0.32g (where 1000mg = 1g)

Thus; n = 0.32/63

n = 0.0051 moles

Volume of acid = 290 mL ⇔ 0.29 L (where 1000 mL = 1L)

Molarity = n/volume;

0.0051/0.29

Molarity = 0.018M

pH = -log 0.018

pH = 1.7

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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