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2 years ago

12

Calculate the ph of a solution that is 0.6 m in sodium acetate and 0.2 m in acetic acid. ka = 1.8 × 10−5 for acetic acid. 1. ph

= 4.3 2. ph = 4.7 3. ph = 5.2 4. ph = 2.7 5. ph = 5.6

1 answer:

Arte-miy333 [17]2 years ago

4 0

0.53 x 200ml = 106 ml of the pH 9.0 buffer + 94 ml of the pH 10 buffer gives the desired solution
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Scene B depicts chemical change in matter at atomic change.

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1 year ago

In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe

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Explanation:

It is given that r = 0.283 nm. As 1 nm = $10^{-9} m$ .

Hence, 0.283 nm = $0.283 \times 10^{-9} m$

Formula for coulombic energy is as follows.

$U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}$

where, e = $1.6 \times 10^{-19}$ C

$\epsilon_{o}$ = $8.85 \times 10^{-12}$

$U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}$

= $1.423 \times 10^{-18} J$

As 1 eV = $1.6 \times 10^{-19} J$

So, 1 J = $\frac{1 eV}{1.6 \times 10^{-19}}$

Hence, U = $\frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}$

= 8.9 eV

Also, 1 J = $\frac{10^{-3} kJ}{6.022 \times 10^{23}mol}$

= $1.67 \times 10^{-27}$ kJ/mol

Therefore, U = $1.423 \times 10^{-18} J \times 1.67 \times 10^{-27}$ kJ/mol

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3 years ago

I need help on chemistry work

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Answer:

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Explanation:

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2 years ago

Which of the following buffers will be most effective at pH 9.25? Group of answer choices a mixture of 1.0 M HC2H3O2 and 1.0 M N

ludmilkaskok [199]

Answer:

The most effective buffer at pH 9.25 will be a mixture of 1.0 M NH3 and 1.0 M NH4Cl

Explanation:

Step 1: Data given

pH of a buffer = pKa + log ([A-]/[Ha])

a mixture of 1.0 M HC2H3O2 and 1.0 M NaC2H3O2 (Ka for acetic acid = 1.8 x 10-5)

pH = -log( 1.8 * 10^-5) + log (1/1)

pH = -log( 1.8 * 10^-5)

pH = 4.74

a mixture of 1.0 M NaCN and 1.0 M KCN (Ka for HCN = 4.9 x 10-10)

pH = -log( 4.9 * 10^-10) + log (1/1)

pH = -log( 1.8 * 10^-5)

pH = 9.30

a mixture of 1.0 M HCl and 1.0 M NaCl

The solution made from NaCl and HCl will NOT act as a buffer.

HCl is a strong acid while NaCl is salt of strong acid and strong base which do not from buffer solutions hence due to HCl PH is less than 7.

a mixture of 1.0 M NH3 and 1.0 M NH4Cl (Kb for ammonia = 1.76 x 10^-5)

Ka * Kb = 1*10^-14

Ka = 10^-14 / 1.76*10^-5

Ka = 5.68*10^-10

pH = -log( 5.68*10^-10) + log (1/1)

pH = -log( 5.68*10^-10)

pH = 9.25

The most effective buffer at pH 9.25 will be a mixture of 1.0 M NH3 and 1.0 M NH4Cl

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3 years ago