A52.0-ml volume of 0.35m ch3cooh (ka= 1.8*10^-5) is titrated with 0.40m naoh . calculate the ph after the addition of 21.0ml of naoh.

Ka= 1,8•10^-5 
pKa = - log(Ka) = 4,74 

Equivalence volume : 

Ca x Va = Cb x Vb 

Vb = Ca x Va / Cb 
Vb = 0,35 x 52,0 / 0,40 
Vb = 45,5 mL 

21,0 mL < 45,5 mL 

. . . then : 

pH = pKa / 2 - log( (Ca x Va - Cb x V) / (Va + V)) / 2 
pH = pKa / 2 - log( (0,35 x 52 - 0,40 x 21) / (52 + 21)) / 2 
pH = 3,24

8.7 degrees celsius and 15.6 degrees fahrenheit

Ithink this is what you meant. the temperature at which a liquid begins to boil is called the boiling point. the boiling point of water is 212°f or 100°c.  and at that point will start to turn into steam (gas)