Grade 11Physical Chemistry
Question no 78 please answer as soon as possible thanks
1 Answer
Arun
Approved Tutor Answer8 Years agoAccording to the given question, HCl + NaOH → NaCl + H₂O
As per the details that are given in the question, we can calculate the following:-
No. of moles of HCl = (40 ml) (0.1 mol/L)
= (0.04 L) (0.1 mol/L) = 0.004 moles
No. of moles of NaOH = (10 ml) (0.45 mol/L)
= (0.01 L) (0.45 mol/L) = 0.0045 moles
Therefore, excess amount of NaOH will be = 0.0045 moles - 0.004 moles
= 0.0005 moles
Thus, the total volume will be = 0.04 L + 0.01 L = 0.05 L
Molarity of OH will be = 0.0005 moles/0.05 L = 0.01 M
pOH = -log [OH]
= -log 0.01
Therefore, pOH = 2
so, pH = 14 - pOH
Thus, pH of the above solution is = 12
As per the details that are given in the question, we can calculate the following:-
No. of moles of HCl = (40 ml) (0.1 mol/L)
= (0.04 L) (0.1 mol/L) = 0.004 moles
No. of moles of NaOH = (10 ml) (0.45 mol/L)
= (0.01 L) (0.45 mol/L) = 0.0045 moles
Therefore, excess amount of NaOH will be = 0.0045 moles - 0.004 moles
= 0.0005 moles
Thus, the total volume will be = 0.04 L + 0.01 L = 0.05 L
Molarity of OH will be = 0.0005 moles/0.05 L = 0.01 M
pOH = -log [OH]
= -log 0.01
Therefore, pOH = 2
so, pH = 14 - pOH
Thus, pH of the above solution is = 12
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