Interpretation:
ka value of given acid is to be calculated.
Concept introduction:
Acid ionization constant tells about equilibrium that whether reactants are favored or products.
For a reaction,
HA+H2O⇌A−+H3O+
Acid ionization constant is given by:
ka=[A−]×[H3O+][HA]where,[A−] is concentration of A−[H3O+]is concentration of H3O+[HA]is concentration of HA
And pH+pOH=14
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Answer to Problem 34PP
ka=6.3×10−4
The acid is HF
Explanation of Solution
Ionization equation is given by:
HX+H2O⇌X−+H3O+
Given data:
Before reaction,[HX]=0.0091 M
pOH=11.32
pH=14−pOH=14−11.32=2.68
Now [X−]=[H+]
pH=−log[H+]
Therefore,
pH=2.68−log[H+]=2.68[H+]=Antilog(−2.68)=2.1×10−3 M
At equilibrium,[HX]=0.0091−2.1×10−3 =0.0070 M
Therefore,
ka=2.1×10−3×2.1×10−30.0070=6.3×10−4
Thus using the table, acid is HF
The value of ka of HF is ka=6.3×10−4
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