The pH of a solution that is 0-1M NaA and 0-1M HA Ka-1 - 10-6 would be-

PH=pka +log [Salt]/[acid]_= log (1× 10^ 6) + log 0.1/0.1= 6 ...

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Ostwald Dilution Law Mathematical Interpretation

The pH of a s...

Question

The $p H$ of a solution that is 0.1M NaA and 0.1M HA $(K a = 1 \times 10^{- 6})$ would be:

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100 mL

0.1 M

H A

N a A

N a O H

(K_{a} of HA = 10^{- 5})

0.1 M

K_{a}

1 \times 10^{- 5}

0.1 M

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K_{a}

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5.6 \times 10^{- 6}

1

H A

10 %

0.01 M

0.1 M

H A

0.05 M

N a A

5.0 m L

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50 m L

0.1 M

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(K a = 1.8 \times 10^{- 5})

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