The \(\mathrm{pH}\) of a solution is 4 . (a) What is the \(\mathrm{H}_{3} \mathrm{O}^{+}\) concentration? (b) What is the \(\mathrm{OH}^{-}\) concentration? (c) Is this solution acidic or basic?

Short Answer

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(a) The \(H_3O^+\) concentration in the solution is \(10^{-4} M\). (b) The \(OH^-\) concentration in the solution is \(10^{-10} M\). (c) Since the \(H_3O^+\) concentration is higher than the \(OH^-\) concentration, the solution is acidic.

Step by step solution

01

(a) Calculating \(H_3O^+\) concentration

To calculate the \(H_3O^+\) concentration, we need to use the formula relating pH and \(H_3O^+\) concentration, which is: \[pH = -\log_{10}(H_3O^+)\] Given that the pH of the solution is 4, we have: \[4 = -\log_{10}(H_3O^+)\] We need to solve this equation for \(H_3O^+\). First, we will take the power of 10 on both sides: \[10^4 = 10^{-\log_{10}(H_3O^+)}\] Now, since \(10^{-\log_{10}(x)}=x\), we have: \[H_3O^+ = 10^{-4}\]
02

(a) Result

Thus, the \(H_3O^+\) concentration in the solution is \(10^{-4} M\).
03

(b) Calculating \(OH^-\) concentration

To calculate the \(OH^-\) concentration, we need to use the ion-product constant of water, which is given by: \[K_w = [H_3O^+] [OH^-]\] The ion-product constant of water, \(K_w\), is equal to \(1.0 \times 10^{-14} M^2\). Given that we know the \(H_3O^+\) concentration, we can find the \(OH^-\) concentration by rearranging the equation and solving for the \(OH^-\) concentration: \[[OH^-] = \frac{K_w}{[H_3O^+]}\] Now, substitute the given values: \[[OH^-] = \frac{1.0 \times 10^{-14} M^2}{10^{-4} M}\]
04

(b) Result

Thus, the \(OH^-\) concentration in the solution is \(10^{-10} M\).
05

(c) Determining if the solution is acidic or basic

To determine if the solution is acidic or basic, we can compare the concentrations of \(H_3O^+\) and \(OH^-\). If the concentration of \(H_3O^+\) is higher than the concentration of \(OH^-\), the solution is acidic. If the concentration of \(OH^-\) is higher than the concentration of \(H_3O^+\), the solution is basic. In this case, we have \( H_3O^+ = 10^{-4} M \) and \( OH^- = 10^{-10} M \). Since the \(H_3O^+\) concentration is higher than the \(OH^-\) concentration, the solution is acidic.

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