Question

X g of naoh is added in 1l of 0.1m of ch3cooh and ph of resulting solution is 5.74 then value of x is

Answer 1

Answer:

Explanation:

ph=5.74

-log h = 5.74

antilog on both sides

antilog -log h = antilog 5.74

1/h = anti log 5+0.6+0.1+0.04

1/h=10^5*4*1.26*1.10

=10^6*5.544

h=10^-6*0.18

n1v1+n2v2/v1+v2 =h

here v1=v2=1

n1+n2/2=h

n1+0.1 /2 =10^-6*0.18

n1+0.1=10^-6*0.36

n1=10^-6*0.36-0.1

n1=m1(mono acidic base)

x/40=10^-6(0.36-10^5)

x=40*10^-6(0.36-10^5)