<span> 52.0ml of 0.35M CH3COOH : 0.052 L(0.35M) = .0182 mol of CH3COOH.
</span>
<span>31.0ml of 0.40M NaOH : .031 L(0.40M) = .0124 mol of NaOH.
</span>
<span>After the reaction, .0124 Mol CH3COO- is generated and .058 mol CH3COOH is left un-reacted. The concentration would be 12.4/V and 5.8/V, respectively. Therefore:
</span>
<span>pH = -log([H+]) = -log(Ka*[CH3COOH]/[CH3COO-]) </span>
<span>= -log(1.8x10^-5*5.8/12.4) = 5.07</span>
It is.
An acid will be strong when its conjugated base is highly stable, and vice-versa.
That can occur for instance through electronic delocalization.
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The eletron pairs repel each other in a tetrahedral shape.
