Determine [oh-], and ph of a 0.28 m aqueous solution of na2co3. kb=2.1 x 10-4

a) [ OH- ] = 7.564 E-3 M

b) pH = 11.88

Explanation:

Na2CO3 ↔ 2Na+  +  CO32-

0.28          2*(0.28)      0.28

⇒ CO32- + H2O ↔ HCO3-  +  OH-

⇒ Kb = 2.1 E-4 = ( [ OH- ] * [ HCO3- ] ) / [ CO32- ]

mass balance:

⇒ M CO32- = [ HCO3- ] + [ CO32- ] = 0.28(1)

⇒ [ CO32- ] = 0.28 - [ HCO3- ](1.1)

charge balance:

⇒ [ Na+ ] = 2*[CO32- ] + [ HCO3- ] + [ OH- ]  = 2*0.28 = 0.56(2)

if (1) * (-2):

⇒ - 0.56 = -2* [ HCO3- ] - 2*[ CO32- ](3)

So (2) + (3):

⇒ 0 = - [ HCO3- ] + [ OH- ]

⇒ [ OH- ] = [ HCO3- ](4)

replacing (4) and (1.1) in Kb:

⇒ Kb = 2.1 E-4 = [ OH- ]² / ( 0.28 - [ OH- ] )

⇒ [ OH- ]² = 5.88 E-5 - 2.1 E-4 [ OH- ]

⇒ [ OH- ]² + 2,1 E-4 [ OH- ] - 5.88 E-5 = 0

⇒ [ OH- ] = 7.564 E-3 M

⇒ pOH = - Log [ OH- ]

⇒ pOH = 2.12

⇒ pH = 14 - pOH

⇒ pH = 11.88

answer is: nuclear fusion produces elements that are heavier than helium.

nuclei combine to form a heavier nucleus, releasing energy.  

for example, fusion of hydrogen atoms at the sun.  

this fusion nuclear reaction is in the inner core of the sun and hydrogen (atomic number 1) is conveted into helium (atomic number 2).  

the inner core is the central region of the sun and all solar energy is produced in the core by nuclear fusion.