Answer:
The molarity of the new solution is 0.72 M
Explanation:
Step 1: Data given
Volume of the original solution = 360 mL =.360 L
Molarity = 0.87 M
We add 75 mL = 0.075 L
Step 2: Calculate moles
Moles = molarity * volume
Moles = 0.87 M * 0.360 L
Moles = 0.3132 moles
Step 3: Calculate new molarity
The number of moles stays constant
Molarity = moles / volume
Molarity = 0.3132 moles / (0.36+0.075)
Molarity = 0.3132 moles / 0.435 L
Molarity = 0.72 M
The molarity of the new solution is 0.72 M
Answer:
![K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BHCO_3%5E-%5D%7D%7B%5BH_2CO_3%5D%7D)
Explanation:
Several rules should be followed to write any equilibrium expression properly. In the context of this problem, we're dealing with an aqueous equilibrium:
- an equilibrium constant is, first of all, a fraction;
- in the numerator of the fraction, we have a product of the concentrations of our products (right-hand side of the equation);
- in the denominator of the fraction, we have a product of the concentrations of our reactants (left-hand side o the equation);
- each concentration should be raised to the power of the coefficient in the balanced chemical equation;
- only aqueous species and gases are included in the equilibrium constant, solids and liquids are omitted.
Following the guidelines, we will omit liquid water and we will include all the other species in the constant. Each coefficient in the balanced equation is '1', so no powers required. Multiply the concentrations of the two products and divide by the concentration of carbonic acid:
![K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BHCO_3%5E-%5D%7D%7B%5BH_2CO_3%5D%7D)