A 75.0-mL sample of 0.0500 M HCN (Ka = 6.2e-10) is titrated with
0.212 M NaOH....
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Chemistry
A 75.0-mL sample of 0.0500 M HCN (Ka = 6.2e-10) is titrated with0.212 M NaOH. What is the [H+] on the solution after 3.0 mL of0.212 M NaOH have been added?
I know the answer is 3.0 x 10^-9 M but I am not sure about thesteps please help
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Given HCN005 M volume 750 mL NaOH 0212 M volume 30 mL Solution Write the balanced reaction between HCN and NaOH HCN aq NaOH aq NaCNaq H2O l Write net ionic equation HCN aq OH aq CN aq H2O l From the reaction we can see that acid
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