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![[H_3O^{+}] =[HCl] \\\\\to [H_3O^{+}] =0.5064 \ m \\\\\to PH =-\log [H_3O^{+}] \\\\](/tpl/images/0929/8717/b345c.png)
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23.1 g of HCl (a strong acid) is added to water to make 1250 mL of solution. Calculate [H3O+] and pH...
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