15.0 mL of 2.0 M NaOH was added to 10 mL of a sulphuric acid solution. Resultant mixture required 4.3 mL of 0.5 M HCl to neutralise the excess base. Thus, (H_{2}*S * O_{4}) in the original mixture is (a) 2.78 M (b) 1.39 M ( d) 0.139 M (c) 0.695 M
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