91影视

First, let's find the ionization constant (Ka) for HNO鈧� (a weak acid). We can use the given pH value to determine the hydrogen ion concentration (\([H^+]\)): \[ \mathrm{pH} = -\log([H^+]), \] Rearranging this equation for \([H^+]\): \[ [H^+] = 10^{-\mathrm{pH}}. \] Plugging in the given pH value of 3.55: \[ [H^+] = 10^{-3.55} \approx 2.818 \times 10^{-4} \mathrm{M}. \] Next, we know that HNO鈧� ionizes according to the following equation: \[ \mathrm{HNO_2} \rightleftharpoons \mathrm{H^+} + \mathrm{NO_2^-}. \] The Ka expression for this reaction is: \[ K_\mathrm{a} = \frac{[\mathrm{H^+}][\mathrm{NO_2^-}]}{[\mathrm{HNO_2}]}. \] Given that the initial concentrations of HNO鈧� and NaNO鈧� are equal (0.50 M), we can set the dissociated concentration of HNO鈧� to 0.5 M - x and the concentration of NO鈧傗伝 to x, so the equation becomes: \[ K_\mathrm{a} = \frac{([H^+] + x)x}{(0.5 - x)}. \]

Now, we will use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the pKa and concentrations of acid and base: \[ \mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\mathrm{base}]}{[\mathrm{acid}]}, \] where base is the concentration of NO鈧傗伝 and acid is the concentration of HNO鈧�. We can rearrange this equation to find the ratio of base concentration to acid concentration: \[ \frac{[\mathrm{base}]}{[\mathrm{acid}]} = 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}}. \]

Let V鈧� be the volume of HNO鈧� and V鈧� be the volume of NaNO鈧�. We can relate the volumes to the concentrations: \[ \frac{[\mathrm{base}]}{[\mathrm{acid}]} = \frac{0.50 \mathrm{M} \times V鈧倉{0.50 \mathrm{M} \times V鈧亇 = \frac{V鈧倉{V鈧亇. \] Now, we can substitute the value derived in Step 2 into this equation: \[ \frac{V鈧倉{V鈧亇 = 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}}, \] where \(\mathrm{pH} = 3.55\) and \(K_\mathrm{a} = 4.5 \times 10^{-4}\). The total volume is 1.00 L, so we have the following equation for V鈧� and V鈧�: \[ V鈧� + V鈧� = 1.00\mathrm{L}. \]

Solve the system of equations established in Steps 3: \[ 1. \; V鈧� + V鈧� = 1.00\mathrm{L}, \] \[ 2. \; \frac{V鈧倉{V鈧亇 = 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}}. \] From equation 2: \[ V鈧� = 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}}V鈧�, \] Substitute this expression for V鈧� into equation 1: \[ V鈧� + 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}}V鈧� = 1.00\mathrm{L}, \] \[ V鈧�(1 + 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}}) = 1.00\mathrm{L}. \] Then, isloate V鈧�: \[ V鈧� = \frac{1.00\mathrm{L}}{1 + 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}}}. \] Now we can plug in the given pH value and calculated pKa: \[ V鈧� \approx 0.723\mathrm{L}. \] Then, we can find V鈧� by subtracting V鈧� from the total volume (1.00 L): \[ V鈧� = 1.00\mathrm{L} - V鈧� \approx 0.277\mathrm{L}. \] Thus, to prepare 1.00 L of a solution buffered at pH 3.55, we need to mix approximately 0.723 L of 0.50 M HNO鈧� and 0.277 L of 0.50 M NaNO鈧�.