A solution has [c6h5cooh] = 0.100 m and [ca(c6h5coo)2] = 0.200 m. ka = 6.3 × 10−5 for c6h5cooh. the solution volume is 5.00 l. what is the ph of the solution after 10.00 ml of 5.00 m naoh is added?

6.8 is the pH of the solution after 10 ml of 5M NaOH is added.

Explanation:

Data given:

Molarity of C6H5CCOH = 0.100 M

molarity of ca(c6h5coo)2  = 0.2 M

Ka = 6.3 x 10^-5

first pH is calculated of the buffer solution

pH = pKa+ log 10

pKa = -log10[Ka]

pka = -log[6.3 x10^-5]

pKa = 4.200

putting the values to know pH of the buffer

pH = 4.200 + log 10

pH = 4.200 + 0.3

    pH  = 4.5 (when NaOH was not added, this is pH of buffer solution)

now the molarity of the solution is calculated after NaOH i.e Mbuffer is added

MbufferVbuffer = Mbase Vbase

putting the values in above equation:

Mbuffer =

             =

             = 0.01 M

molarity or [ A-] = 5M

pH =   pKa+ log 10

pH = 4.200 + log 10

pH = 4.200+ 2.69

pH = 6.8

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml