If 10.0 mmol of sodium hydroxide is added to 1.00 L of a buffer that is 0.200 M HClO(aq)and 0.155 M NaHClO(aq),the final concentrations of HClO(aq)and ClO^-(aq),respectively,are

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The addition of NaOH will react with HClO, decreasing its concentration and increasing the concentration of ClO⁻. The reaction is: HClO + OH⁻ → ClO⁻ + H₂O. The initial moles of HClO are 0.200 mol/L * 1.00 L = 0.200 mol, and NaOH added is 0.010 mol. After the reaction, HClO will be 0.200 mol - 0.010 mol = 0.190 mol, and ClO⁻ will be 0.155 mol + 0.010 mol = 0.165 mol. Dividing by 1.00 L gives concentrations of 0.190 M HClO and 0.165 M ClO⁻.

If 100 Mmol of Sodium Hydroxide Is Added to 1