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coldgirl [10]
1 year ago
9

What is the ph of a solution of 1.699 l of 1.25 m hcn, ka = 6.2 x 10-10, and 1.37 moles of nacn?

Chemistry
1 answer:
BlackZzzverrR [31]1 year ago
8 0

The pH of a solution is 9.02.

c(HCN) = 1.25 M; concentration of the cyanide acid

n(NaCN) = 1.37 mol; amount of the salt

V = 1.699 l; volume of the solution

c(NaCN) = 1.37 mol ÷ 1.699 l

c(NaCN) = 0.806 M; concentration of the salt

Ka = 6.2 × 10⁻¹⁰; acid constant

pKa = -logKa

pKa = - log (6.2 × 10⁻¹⁰)

pKa = 9.21

Henderson–Hasselbalch equation for the buffer solution:

pH = pKa + log(cs/ck)

pH = pKa + log(cs/ck)

pH = 9.21 + log (0.806M/1.25M)

pH = 9.21 - 0.19

pH = 9.02; potential of hydrogen

More about buffer: brainly.com/question/4177791

#SPJ4

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How many equivalents are present in 10 g of Ca2+? <br> 1 <br> 0.5 <br> 1.5 <br> 2
Lynna [10]

0.5

Explanation:

Given parameters:

Mass of Ca²⁺ = 10g

unknown:

Equivalent weight = ?

Solution:

Equivalent weight that is the amount of electrons which a substance gains or loses per mole.

Ca²⁺ has +3 charge

It lost 2e⁻;

therefore;

  In 1 mole of  Ca²⁺, we have 2 equivalent weight

1 mol  Ca²⁺ = 2eq. wts. 

1 mol Ca x (40 g / 1 mol ) x (1 mol / 2 eq.wts.) = 20.0 g = 1 eq.wt. 

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10.0 g  Ca²⁺ x (1 eq.wt. / 20.0 g) = 0.5 eq.wts.

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8 0
3 years ago
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List the protons, neutrons, and electrons for Ge2+<br> Please hurry
Nady [450]

Answer:

32, 30 and 41

Explanation:

The problem here is to find the number of:

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In this ion,

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7 0
3 years ago
. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are
olga nikolaevna [1]

<u>Answer:</u> The standard free energy change of formation of S^{2-}(aq.) is 92.094 kJ/mol

<u>Explanation:</u>

We are given:

K_{sp}\text{ of }Ag_2S=8\times 10^{-51}

Relation between standard Gibbs free energy and equilibrium constant follows:

\Delta G^o=-RT\ln K

where,

\Delta G^o = standard Gibbs free energy = ?

R = Gas constant = 8.314J/K mol

T = temperature = 25^oC=[273+25]K=298K

K = equilibrium constant or solubility product = 8\times 10^{-51}

Putting values in above equation, we get:

\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ

For the given chemical equation:

Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)

The equation used to calculate Gibbs free change is of a reaction is:  

\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]

The equation for the Gibbs free energy change of the above reaction is:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]

We are given:

\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ

Putting values in above equation, we get:

285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol

Hence, the standard free energy change of formation of S^{2-}(aq.) is 92.094 kJ/mol

8 0
3 years ago
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