See the attachment below. remember, i will report any incorrect answers.

H2CO3 + H2O ⇄ H3O+  + HCO-

.18                              0            0

-x                               +x             +x

.18-x                           x                 x

x^2 ÷ .18 = Ka1 = 4.3 X 10^ -7

x= 2.78 X 10^ -4

HCO- + H2O ⇄ H3O+  + CO-

2.78 X 10^ -4          2.78 X 10^ -4                0

-x                               +x             +x

2.78 X 10^ -4 -x           2.78 X 10^ -4+ x          x

k a2 = 5.6 X 10^ -11 = (2.78 X 10^ -4 + x)(x) ÷ (2.78 X 10^ -4 - x)

x= 5.6 X 10^ -11  

Total [ H3 O +] = 2.78 X 10^ -4  + 5.6 X 10^ -11  = 2.78 X 10^ -4

pH = - log (2.78 X 10^ -4 )= 3.555 ~ 3.56

The answer is B

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong