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27.0 ml of 0.45 m naoh is neutralized by 20 ml of an hcl solution the molarity of the hcl solution is?
Wiki User
∙ 13y ago
Given: 27 mL of NaOH, 0.45M;
20 mL HCI
Need: M of HCI
27 ml NaOH*(1 L NaOH/1000mL NaOH)*(0.45M NaOH/1L NaOH)*(1mole HCI/1 mole NaOH)=0.012
0.012/0.02=0.607 M HCI (or rounded 0.61 M HCI)
Wiki User
∙ 13y ago
The number of moles of NaOH in the solution can be calculated as (0.027 L) * (0.45 mol/L) = 0.01215 mol. Since the reaction is 1:1 between NaOH and HCl, 0.01215 mol of HCl was neutralized. The molarity of the HCl solution is (0.01215 mol) / (0.020 L) = 0.6075 M.
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