IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

ayshwarya

1)25 mL of a saturated solution of phenyl acetic acid requires 17.70 mL of 0.1850 M NaOH for its neutralisation. What is the pH of saturated solution of phenyl acetic acod . Assume Ka for the acid = 5.56 x 10^-5 (ans pH = 2.57 )

2)Calculate the pH of a solution obtained by mixing 0.1 ltr of a strong acid solution of pH = 4, and o.2 ltr a solution of strong base of pH = 10. ( ans 9.5228)

3) calculate the pH of a solutin made by mixing 50 mL of 0.2 M NH4Cl and 75 mL of 0.1 M NaOH Given Kb for NH4 OH = 1.8 x 10 ^-5 (Ans 9.74 )

4) Calculate approximate pH of 0.05 M solution of carbonic acid . Given

H2 CO3

H + +HCO3^- : K1 = 4.3 x 10^-7

HCO3^-

H ^+ + CO3^-2 : K2 = 5.6 x 10 ^-11

5) calculate the pH of a solution obtained by mixing 60 mL of 0.45 N NaOH and 40 mL of 0.3 N HCl . (Ans 13.17 )

6) 0.12 mL of a solution of KOH ( 50% by weight of KOH) , specific gravity 1.5 g/cm^3 is diluted to 250 mL . calculate its pH. (ans 11.80 )

7) calculate pH of a solution of benzoic acid in water being soluble upto 2.06 g/dm^3.Ka=6.4x10^-5 ( ans 3 )

8) A weak base BOH of concentration 0.02 mol /ltr has a pH if 10.45 , If 100 mL of this base are mixed with 10 mL of 0.1 M HCl, what will be the pH of mixture. (ans 8.598 ).

9) At _{^{25 degree C what changes in pH occur when o.1 millimole of a strong acid is added to 25 mL of : (a) H2O (b) 0.025 M HCl (c) 0.025 M Na OH ? ( Ans (a) 4.602 (b) 0.064 (c) 0.076 )}}

LAMPARD

4>Solved it here...
http://www.goiit.com/posts/list/physical-chemistry-ionic-equilibrium-47925.htm#239785

LAMPARD

5>Solved it here..
http://www.goiit.com/posts/list/physical-chemistry-ionic-equilibrium-47926.htm#239258

LAMPARD

1>Let molarity of acid be M.
So,since millieq. are equal,
M*25=17.7*0.185
So,M=0.13098.
Now,u know that for weak acid,a=root(Ka/C) where a is degree of dissociation.
Ka is given and C is what we have calculated as M.
So,a comes out to be 0.021.
Now,the conc. of H+ will be C*a.
So,Ph=-log[C*a]=2.57

LAMPARD

3>Meq of NH₄Cl=10
Meq of NaOH=7.5
So,Meq. of salt formed=10-7.5=2.5
Meq. of base NaOH=7.5
So,Poh=Pkb+log[Meq of salt/Meq of acid]
Subst. values of Kb and meq. of salt and acid to get Poh and then use Ph+Poh=14 to find Ph..

ayshwarya

excuse me mr. lampard can u explain ans. for 4,5ques plzzzzzz...............

eistien

for question 5

Me.eq of NaOH=27

Me.eq of HCl=12
Me.eq of NaOH in resultant=15
Normality of NaOH=15/100
Molarity=15/100
[OH-]=15/100
So,Ph=14-(-log[OH-])

eistien

question 4:

hey K2<<<K1 so contribution due to K2 can be neglected and then

pH=-(1/2)log((K1)c)

where c is the concentration !!!!

am i rite???

LAMPARD

8>Poh=3.55
So,[OH-]=2.82*10^-4.
So,c*a=2.82*10^-4. where c is conc.=0.02 and a is degree of diss.
So,a=0.0141.
But,ca²=Kb ,where Kb is dissociation constant of BOH.
Thus,Kb=3.97*10^-6.
Now,during expt.,Meq of acid=1
Meq of base=2.
Thus,after neutralization,Meq. of salt formed=1 and meq of base left=1
Now,using Poh=PKb+log(meq of salt/meq of base),
you get Poh=5.401.
Thus Ph=14-5.401=8.598.

karthik2007

Qn 2)

Number of moles of H⁺ = 10^-10

Number of moles of OH^- = 2 x 10^-5

Now, as the number of H⁺ ions is less, it will completely react with the available OH^- ions to form water, and the resulting Hydroxide ions contribute to the pOH.

Hence, number of moles of OH^- remaining = 1.99 x 10^-5

Hence, pOH = 4.6989

Hence, pH = 14-pOH = 9.301

sagar90

hey karthik how did u found pOH without using log table can u please tell me?

karthik2007

well, 1.99 can be taken as 2, and dont u know the log of 2?

By the way, I used a calculator for this problem

sagar90

one more thing yaar
how did u find h+ ion and OH- concentration?

ananth_patri

pH is given for each no?? so pH =-logH+....hope u got it....

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