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How many milliliters of 0.103 M HCl would be required to titrate 6.46 g KOH?
Wiki User
∙ 14y ago
6.46g(KOH) / 56.108 g/mol(KOH) = 0.1151 mol(KOH) = 0.1151 mol(OH-) , which needs of the titrant the same amount of acid, so:
0.1151 mol H+ = 0.1151 mol(HCl) / 0.103 mol(HCl)/L(titrant) = 1.118 L of this titrant ( 0.103 M HCl)
= 1118 ml
Wiki User
∙ 14y ago
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