The below given compound is slightly soluble in water.
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Answer:
Explanation:
We are given that 25 mL of 0.10 M is titrated with 0.10 M NaOH(aq).
We have to find the pH of solution
Volume of
Volume of NaoH=0.01 L
Volume of solution =25 +10=35 mL=
Because 1 L=1000 mL
Molarity of NaOH=Concentration OH-=0.10M
Concentration of H+= Molarity of =0.10 M
Number of moles of H+=Molarity multiply by volume of given acid
Number of moles of H+==0.0025 moles
Number of moles of =0.001mole
Number of moles of H+ remaining after adding 10 mL base = 0.0025-0.001=0.0015 moles
Concentration of H+=
pH=-log [H+]=-log [4.28]=-log4.28+2 log 10=-0.631+2