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labwork [276]
3 years ago
10

A 25. 0-ml sample of 0. 150 m hydrocyanic acid is titrated with a 0. 150 m naoh solution. What is the ph beforeany base is added

? the ka of hydrocyanic acid is 4. 9 × 10-10.
Chemistry
1 answer:
Ray Of Light [21]3 years ago
4 0

Answer:

stars,auroras,a neon sign

Explanation:

just did it

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A 1.0l buffer solution contains 0.100 mol of hc2h3o2 and 0.100 mol of nac2h3o2. the value of ka for hc2h3o2 is 1.8×10−5. part a
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There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as

pH = pKa + log [A]/[HA] 

 where pKa = - log [Ka]
[A] is the concentration of the conjugate base
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Given:
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Solution:
pKa = - log ( 1.8x10^-5) = 4.74

[A] = 0.015 mol + 0.100 mol = .115 moles
[HA] = .1 - 0.015 = 0.085 moles

pH = 4.74 + log (.115/0.085)
pH = 4.87
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