Question

A 20.00 mL solution of 0.095 M NaI is titrated with 0.062 M AgNO$_3$. Given the $K_{sp}$ of AgI is 8.3\times10^{-17}. When 55.2 mL of AgNO$_3$ is added, calculate pAg$^+$ of the final solution.

          A 20.00 mL solution of 0.095 M NaI is titrated with
0.062 M AgNO$_3$. Given the $K_{sp}$ of AgI is 8.3\times10^{-17}.
When 55.2 mL of AgNO$_3$ is added, calculate pAg$^+$ of
the final solution.

A 20.00 mL solution of 0.095 M NaI is titrated with
0.062 M AgNO3. Given the Ksp of AgI is 8.3×10^-17.
When 55.2 mL of AgNO3 is added, calculate pAg^+ of
the final solution.

Added by Pedro H.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert David Collins

07/11/2023

Step 1

Moles of Nal = (0.095 mol/L) * (20.00 mL / 1000 mL) = 0.0019 mol Show more…

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A 20.00mL solution of 0.095MNal is titrated with 0.062MAgNO_(3). Given the K_(sp) of Agl is 8.3 imes 10^(-17). When 55.2mL of AgNO_(3) is added, calculate pAg^(+)of the final solution. The answer 14.38 was wrong A 20.00 mL solution of 0.095 M Nal is titrated with When 55.2 mL of AgNO3 is added,calculate pAg* of the final solution. 14.38