Consider the following reaction: CH3COOH + H2O CH3COO- + H3O+
A buffer solution is prepared by adding NaCH3COO(s) to CH3COOH(aq). What happens to the equilibrium when a few drops of NaOH solution are added to the buffer?

A. It shifts left and [CH3COO-] increases.
B. It shifts left and [CH3COO-] decreases.
C. It shifts right and [CH3COO-] increases.
D. It shifts right and [CH3COO-] decreases.

I would say:

C. It shifts right and [CH3COO-] increases.

Explanation:

I am uncertain, but since NaOH is basic I think it should shift right.

hope this helps though

the volume of fluid displaced would equal the volume of the gas. use a gas syringe - this is a laboratory instrument that can be used to withdraw a certain volume of gas from a closed system or to measure the volume of gas from a chemical reaction.

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"