Question

Complete the following table by calculating the missing entries. In each case indicate whether the solution is acidic or basic. $$ \begin{array}{lllll} \hline & & & & \text { Acidic or } \\ \mathrm{pH} & \mathrm{pOH} & {\left[\mathrm{H}^{+}\right]} & {\left[\mathrm{OH}^{-}\right]} & \text {basic? } \\ \hline 11.25 & & & & \\ & 6.02 & & \\ & & 4.4 \times 10^{-4} \mathrm{M} & & \\ & & & 8.5 \times 10^{-3} \mathrm{M} & \\ \hline \end{array} $$

Short answer

} \\\ \hline 11.25 & 2.75 & 5.62 \times 10^{-12} M & 1.78 \times 10^{-3} M & \text { basic } \\\ 7.98 & 6.02 & 1.05 \times 10^{-8} M & 9.33 \times 10^{-7} M & \text { basic } \\\ 3.36 & 10.64 & 4.4 \times 10^{-4} M & 2.27 \times 10^{-11} M & \text { acidic } \\\ 11.93 & 2.07 & 1.18 \times 10^{-12} M & 8.5 \times 10^{-3} M & \text { basic } \\\ \hline \end{array} $$

Step-by-step solution

First Row: Finding pOH, [H+] and [OH-]

Given: pH = 11.25 Using relationship 3, pOH = 14 - pH = 14 - 11.25 = 2.75 Now, we will find [H+] and [OH-] values using relationships 1 and 2: [H+] = 10^(-pH) = 10^(-11.25) = 5.62 × 10^(-12) M [OH-] = 10^(-pOH) = 10^(-2.75) = 1.78 × 10^(-3) M Since pH > 7, the solution is basic.

Second Row: Finding pH, [H+] and [OH-]

Given: pOH = 6.02 Using relationship 3, pH = 14 - pOH = 14 - 6.02 = 7.98 Now, we will find [H+] and [OH-] values using relationships 1 and 2: [H+] = 10^(-pH) = 10^(-7.98) = 1.05 × 10^(-8) M [OH-] = 10^(-pOH) = 10^(-6.02) = 9.33 × 10^(-7) M Since pH > 7, the solution is basic.

Third Row: Finding pH, pOH and [OH-]

Given: [H+] = 4.4 × 10^(-4) M Using relationship 1, pH = -log[H+] = -log(4.4 × 10^(-4)) = 3.36 Using relationship 3, pOH = 14 - pH = 14 - 3.36 = 10.64 Now, we will find the [OH-] value using relationship 4: [OH-] = (1 × 10^(-14) M²) / [H+] = (1 × 10^(-14) M²) / (4.4 × 10^(-4) M) = 2.27 × 10^(-11) M Since pH < 7, the solution is acidic.

Fourth Row: Finding pH, pOH and [H+]

Given: [OH-] = 8.5 × 10^(-3) M Using relationship 2, pOH = -log[OH-] = -log(8.5 × 10^(-3)) = 2.07 Using relationship 3, pH = 14 - pOH = 14 - 2.07 = 11.93 Now, we will find the [H+] value using relationship 4: [H+] = (1 × 10^(-14) M²) / [OH-] = (1 × 10^(-14) M²) / (8.5 × 10^(-3) M) = 1.18 × 10^(-12) M Since pH > 7, the solution is basic. The completed table is: $$ \begin{array}{lllll} \hline & & & & \text { Acidic or } \\\ \mathrm{pH} & \mathrm{pOH} & {\left[\mathrm{H}^{+}\right]} & {\left[\mathrm{OH}^{-}\right]} & \text {basic? } \\\ \hline 11.25 & 2.75 & 5.62 \times 10^{-12} M & 1.78 \times 10^{-3} M & \text { basic } \\\ 7.98 & 6.02 & 1.05 \times 10^{-8} M & 9.33 \times 10^{-7} M & \text { basic } \\\ 3.36 & 10.64 & 4.4 \times 10^{-4} M & 2.27 \times 10^{-11} M & \text { acidic } \\\ 11.93 & 2.07 & 1.18 \times 10^{-12} M & 8.5 \times 10^{-3} M & \text { basic } \\\ \hline \end{array} $$