Question

0.1M, 100ml,h2so4 is mixed with 0.2N, 900mlNaOH. Find the ph of the solution. ​

Answer 1

QUESTION ⤵️

0.1m, 100ml, H2SO4 is mixed with 0.2N, 900 ml NaOH. find the pH of solution??

ANSWER⤵️

H2SO4 ➡➡ 0.1M , 100Ml

NaOH ➡➡0.2N, 900Ml

in H2SO4 we can write 0.2N because molarity= normality ×valency factor.and valency factor is 1.so, N=M

NaVa=0.2/10×100

${\huge{\boxed{\mathcal{\red{NaVa\:=\:20}}}}}$

NbVb= 0.2/10×900

${\huge{\boxed{\mathcal{\pink{NbVb\:=\:180}}}}}$

NbVb>NaVa

{OH-} resultant= NbVb-NaVa/1000

= 180-20/1000

${\huge{\boxed{\mathcal{\blue{OH-\: =\: 16× 1o^-2}}}}}$

pOh = 2-log16

= 2-log4^2

= 2-2log4

= 2-2(0.60)

${\huge{\boxed{\mathcal{\orange{pOH\:=\: 0.8}}}}}$

then we can easily find pH

pH = 14-0.8

${\huge{\boxed{\mathcal{\orange{pH\:=\: 13.2}}}}}$

$\huge\mathfrak\purple{hope\:it\:hope}$

Answer 2

Answer:

A mixture containing 100 ml of 0.1 M acetic acid and 50 ml of 0.1 M NaOH is acidic buffer soluion. Its pH is given by the expression pH=pK  

a

​  

+log  

acid

salt

​  

.

Substitute values in the above expression.

5=pK  

a

​  

+log  

0.1×0.05

0.1×0.05

​  

 or pK  

a

​  

=5.

When 100 ml of 0.05 M NaOH  is added, the acid is completely neutralized and the solution contains salt sodium acetate.

The expression for the hydrogen ion concentration of the salt of weak acid and strong base is pH=  

2

1

​  

(pK  

w

​  

+pK  

a

​  

+logc).

Substitute values in the above expression.

pH=  

2

1

​  

(14+5+log(  

0.25

0.1

​  

))=8.8.

Hence, the change in pH is ΔpH=8.8−5=3.8.

Explanation: