What is the pH of a 0.055 M solution of sodium cyanide? The K_a value for hydrocyanic acid is 6.2 $\times$ 10^-10.

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The pH of a 0.055 M solution of sodium cyanide can be calculated using the formula for the dissociation of weak acids, given that sodium cyanide (NaCN) dissociates in water to produce cyanide ions (CN-) which then react with water to produce hydrocyanic acid (HCN) and hydroxide ions (OH-). The $K_a$ value for hydrocyanic acid (HCN) is given as $6.2 \times 10^{-10}$. Since sodium cyanide is a salt of this weak acid, we use the $K_a$ value to find the $K_b$ for CN- because $K_w = K_a \times K_b$, where $K_w = 1.0 \times 10^{-14}$ at 25°C. $$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{6.2 \times 10^{-10}} = 1.61 \times 10^{-5}$$Using the $K_b$ and the concentration of CN- (which is the same as the initial concentration of NaCN, 0.055 M), we can find the $pOH$ of the solution using the formula for $K_b$:$$K_b = \frac{[OH^-][HB^+]}{[B]}$$Assuming $x$ is the concentration of OH- produced, $$1.61 \times 10^{-5} = \frac{x^2}{0.055 - x} \approx \frac{x^2}{0.055}$$Solving for $x$ gives the concentration of OH-. Then, $pOH = -\log[OH^-]$ and $pH = 14 - pOH$. Given the calculations and the options provided, the correct pH is closest to 10.97, indicating that the solution is basic, which aligns with the expected behavior of a cyanide solution.

What Is the PH of a 0 ×\times× 10-10

What Is the PH of a 0 $\times$ 10^-10