Chemistry, 01.08.2019 06:10 taryn1932

You need to prepare 100.0 ml of a ph 4.00 buffer solution using 0.100 m benzoic acid (=4.20) and 0.240 m sodium benzoate. how many milliliters of each solution should be mixed to prepare this buffer?

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Answer from: whitneyb13

Explanation:

As it is given that $pK_{a}$ = 4.20 and pH is 4.0.

Now, we use Henderson-Hasslbach equation as follows.

pH = $pK_{a} + log \frac{[salt]}{acid}$

So, it is given that salt is sodium benzoate and acid is benzoic acid. Hence, the given equation will become as follows.

pH = $pK_{a} + log \frac{[\text{sodium benzoate}]}{\text{[benzoic acid]}}$ ........ (1)

Now, substitute the given values into equation (1) as follows.

pH = $pK_{a} + log \frac{[\text{sodium benzoate}]}{\text{[benzoic acid]}}$

4.0 = $4.20 + log \frac{[{C_{6}H_{5}COONa}]}{[C_{6}H_{5}COOH]}$

$log \frac{[{C_{6}H_{5}COONa}]}{[C_{6}H_{5}COOH]}$ = 4.0 - 4.20

= - 0.20

$\frac{[{C_{6}H_{5}COONa}]}{[C_{6}H_{5}COOH]}$ = $10^{-0.20}$

= 0.613

or, $[C_{6}H_{5}COONa]$ = $0.613 \times [C_{6}H_{5}COOH]$ ............. (2)

Since, the total volume of (acid + base) is given as 100.0 mL. So, let us assume volume of acid is x and volume of base is y.

Hence, x + y = 100 mL ....... (3)

And, equation (2) will become as follows.

$y \times 0.240 M$ = $0.613 \times x \times 0.1 M$

y = 0.255x

Substituting value of y into equation (3) as follows.

x + 0.255x = 100 mL

1.255 x = 100 mL

x = 79.68 mL

So, value of y will be as follows.

y = 0.255 x

= 0.255 × 79.68 mL

= 20.31 mL

Thus, we can conclude that volume of benzoic acid is 79.68 mL and volume of sodium benzoate is 20.31 mL.

Answer from: Quest

remember that hf is only the energy of one photon. not two. not three. one. 1) 5.12 * 10^3 photons find the energy of one photon, then determine how many photons are needed to reach the sensitivity. e = hf = (6.626 * 10^-34 js)(5.95 * 10^19 hz) = 3.94247 * 10^-14 j per photon (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons round to three sig figs. 2) 5.77 * 10^-17 j energy of pulse = energy per photon * number of photons = hfn = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250) = 5.774559 * 10^-17 j round to three sig figs. 3) 4.07 * 10^8 kj/mol 10^12 pm = 1 m c = fλ f = c/λ e = hf = hc/λ = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol) = 40702571.4 kj/mol round to three sig figs. 4) 9.94 * 10^-1 m e = hc/λ λ = hc / e = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j) = 0.9939 m round to three sig figs. 5) 9.06 * 10^3 photons energy of pulse = energy per photon * number of photons number of photons = energy of pulse / energy per photon = energy of pulse / (hf) = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)] = 9055.23695 photons round to three sig figs. 6) 2.78 * 10^-15 j this is problem 1, except in reverse. sensitivity = minumum number of photons * energy per photon = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz) = 2.78070758 × 10^-15 j round to three sig figs.

Answer from: Quest

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You need to prepare 100.0 ml of a ph 4.00 buffer solution using 0.100 m benzoic acid (=4.20) and 0.2...

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