Calculate the ph of the solutions obtained by mixing:

a) 50.0 cm3 of 0.25 m ch3cooh and 62.5 cm3 of 0.20 m nach3coo

b) 40.0 cm3 of 0.5 m nh4oh and60 cm3 of 1.0 m nh4cl

c) v cm3 of 0.1 m nah2po4 and 3v cm3 of 0.1 m na2hpo4 (it is actually v not a typo)

ka (ch3cooh) = 1.73 x 10^-5 ; kb (nh3) = 1.8 x 10^-5 ; ka2 (h3po4) = 6.2 x 10^-8

the millers tale is the answer.

"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"

i would say b

explanation:

but i think i would need more information than that