WEB To make a buffer with \(\mathrm{pH}=3.0\) from \(\mathrm{HCHO}_{2}\) and \(\mathrm{CHO}_{2}^{-}\), (a) what must the [HCHO \(\left._{2}\right] /\left[\mathrm{CHO}_{2}^{-}\right]\) ratio be? (b) how many moles of \(\mathrm{HCHO}_{2}\) must be added to a liter of \(0.139 \mathrm{M}\) \(\mathrm{NaCHO}_{2}\) to give this \(\mathrm{pH}\) ? (c) how many grams of \(\mathrm{NaCHO}_{2}\) must be added to \(350.0 \mathrm{~mL}\) of \(0.159 \mathrm{MHCHO}_{2}\) to give this \(\mathrm{pH}\) ? (d) What volume of \(0.236 \mathrm{M} \mathrm{HCHO}_{2}\) must be added to \(1.00 \mathrm{~L}\) of a \(0.500 \mathrm{M}\) solution of \(\mathrm{NaCHO}_{2}\) to give this pH? (Assume that volumes are additive.)

Step by step solution

01

Write down the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is given by: \(p\mathrm{H} = p\mathrm{K}_{a} + \log \frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\) where \(\mathrm{pH}\) is the pH of the buffer, \(p\mathrm{K}_{a}\) is the \(p\mathrm{K}_{a}\) value of the weak acid, \([\mathrm{A}^{-}]\) is the concentration of the conjugate base, and \([\mathrm{HA}]\) is the concentration of the weak acid.

02

Calculate the pKa of the weak acid

The weak acid is \(\mathrm{HCHO}_{2}\) (formic acid), which has a \(p\mathrm{K}_{a}\) the value of 3.75.

03

Find the ratio of acid to base concentrations

Using the Henderson-Hasselbalch equation and the given pH (3.0), we can find the ratio of acid to base concentrations: \(3.0 = 3.75 + \log \frac{[\mathrm{CHO}_{2}^{-}]}{[\mathrm{HCHO}_{2}]}\) Solve this equation for the ratio of acid to base concentrations: \(\log \frac{[\mathrm{CHO}_{2}^{-}]}{[\mathrm{HCHO}_{2}]} = 3.0 - 3.75\) \(\frac{[\mathrm{CHO}_{2}^{-}]}{[\mathrm{HCHO}_{2}]} = 10^{-0.75}\) The ratio is approximately \(0.1778\). This means that the concentration of the conjugate base ([CHO\(_2^{-}\)]) must be about 0.1778 times the concentration of the weak acid ([HCHO\(_2\)]) to achieve a pH of 3.0. Now, we can use this ratio to solve the remaining parts of the exercise:

04

Moles of HCHO\(_2\) needed for a 1-L solution of 0.139 M NaCHO\(_2\)

We are given the concentration of NaCHO\(_2\), which corresponds to the concentration of the conjugate base, [CHO\(_2^{-}\)]. Using the ratio, we can find the concentration of the weak acid: \([\mathrm{HCHO}_{2}] = \frac{[\mathrm{CHO}_{2}^{-}]}{0.1778} = \frac{0.139}{0.1778} = 0.781 \mathrm{M}\) Now, multiply this concentration by the volume to obtain the number of moles of HCHO\(_2\) needed: Moles of \(\mathrm{HCHO}_{2} = 0.781 \mathrm{M} \times 1.00 \mathrm{L} = 0.781 \mathrm{mol}\)

05

Grams of NaCHO\(_2\) needed for 350 mL of 0.159 M HCHO\(_2\)

First, find the moles of HCHO\(_2\) in 350 mL of a 0.159 M solution: Moles of \(\mathrm{HCHO}_{2} = 0.159 \mathrm{M} \times 0.350 \mathrm{L} = 0.0557 \mathrm{mol}\) Now, using the ratio, find the moles of NaCHO\(_2\) (which corresponds to moles of CHO\(_2^{-}\)): Moles of \(\mathrm{NaCHO}_{2} = 0.0557 \mathrm{mol} \times 0.1778 = 0.00992 \mathrm{mol}\) Finally, convert the moles of NaCHO\(_2\) to grams: Grams of \(\mathrm{NaCHO}_{2} = 0.00992 \mathrm{mol} \times (23.0 + 12.01 + 16.0 \mathrm{~g/mol}) = 0.685 \mathrm{g}\)

06

Volume of 0.236 M HCHO\(_2\) needed for 1 L of 0.500 M NaCHO\(_2\)

First, find the moles of NaCHO\(_2\) in 1 L of 0.500 M solution: Moles of \(\mathrm{NaCHO}_{2} = 0.500 \mathrm{M} \times 1.00 \mathrm{L} = 0.500 \mathrm{mol}\) Now, using the ratio, find the moles of HCHO\(_2\) needed: Moles of \(\mathrm{HCHO}_{2} = 0.500 \mathrm{mol} \times \frac{1}{0.1778} = 2.81 \mathrm{mol}\) Finally, find the volume of the 0.236 M HCHO\(_2\) solution needed to have 2.81 moles of HCHO\(_2\): Volume of \(\mathrm{HCHO}_{2} \mathrm{~solution} = \frac{2.81 \mathrm{mol}}{0.236 \mathrm{M}} = 11.90 \mathrm{L}\)

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