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Chapter 13: Problem 64

What are the major species present in 0.250 \(M\) solutions of each of the following acids? Calculate the \(\mathrm{pH}\) of each of these solutions. a. \(\mathrm{HOC}_{6} \mathrm{H}_{5}\) b. HCN

Short Answer

Expert verified
The major species in the 0.250 M solution of \(\mathrm{HOC}_{6} \mathrm{H}_{5}\) are \(\mathrm{HOC}_{6} \mathrm{H}_{5}\), \(\mathrm{H}^{+}\), \(\mathrm{OC}_{6} \mathrm{H}_{5}^-\), and the pH is approximately 0.60. For the 0.250 M solution of HCN, the major species are HCN, \(\mathrm{H}^+\), \(\mathrm{CN}^-\), and the pH is approximately 0.60.

Step by step solution

01

Write the chemical equation for ionization

When an acid reacts with water, it donates an H+ ion to the water molecule to form a hydronium ion (H3O+). The general equation is: \[ \mathrm{HA} \rightleftharpoons \mathrm{H^+} + \mathrm{A^-} \] For our problem: \[ \mathrm{HOC}_{6} \mathrm{H}_{5} (aq) \rightleftharpoons \mathrm{H}^+ (aq) + \mathrm{OC}_{6} \mathrm{H}_{5}^- (aq) \]
02

Identify the acid dissociation constant (Ka)

Referring to the available reference data, we can find that in case of \(\mathrm{HOC}_{6}\mathrm{H}_{5}\) we have the acid dissociation constant \(K_a=1.6\times 10^{-5}\).
03

Calculate the equilibrium concentrations using the ICE table or approximation method

Since the dissociation constant \(K_a\) is quite small, we can use the approximation method by assuming the change in the concentration of the major species is negligible. Thus, the concentration of H+ will be approximately equal to the concentration of \(\mathrm{HOC}_{6} \mathrm{H}_{5}\), which is \(0.250\) \(\mathrm{M}\).
04

Calculate the pH of the solution

The pH is determined using the formula pH = \(- \log \left [\mathrm{H^+} \right]\). So, we have: \[ pH = -\log\left(0.250\right) \] \[ pH \approx 0.60 \] The major species in the 0.250 M solution of \(\mathrm{HOC}_{6} \mathrm{H}_{5}\) are \(\mathrm{HOC}_{6} \mathrm{H}_{5}\), \(\mathrm{H}^{+}\), \(\mathrm{OC}_{6} \mathrm{H}_{5}^-\), and the pH is approximately 0.60. #b. HCN#
05

Write the chemical equation for ionization

The equation for the ionization of HCN is: \[ \mathrm{HCN} (aq) \rightleftharpoons \mathrm{H}^+ (aq) + \mathrm{CN}^- (aq) \]
06

Identify the acid dissociation constant (Ka)

Referring to the available reference data, we can find that for HCN, we have the acid dissociation constant \(K_a=4.9\times 10^{-10}\).
07

Calculate the equilibrium concentrations using the ICE table or approximation method

Since the dissociation constant \(K_a\) is very small, we can use the approximation method by assuming the change in the concentration of the major species is negligible. Thus, the concentration of H+ will be approximately equal to the concentration of HCN, which is \(0.250\) \(\mathrm{M}\).
08

Calculate the pH of the solution

The pH is determined using the formula pH = \(- \log \left [\mathrm{H^+} \right]\). So, we have: \[ pH = -\log\left(0.250\right) \] \[ pH \approx 0.60 \] The major species in the 0.250 M solution of HCN are HCN, \(\mathrm{H}^+\), \(\mathrm{CN}^-\), and the pH is approximately 0.60.

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Most popular questions from this chapter

A solution is prepared by dissolving 0.56 g benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}, K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) in enough water to make \(1.0 \mathrm{L}\) of solution. Calculate \(\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right],\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}\right],\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right]\) and the pH of this solution.

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Is an aqueous solution of NaHSO_ acidic, basic, or neutral? What reaction occurs with water? Calculate the pH of a 0.10-M solution of \(\mathrm{NaHSO}_{4}\).

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