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3 years ago

As the PH scale decreases or increases, how does this affect the acidity?

Chemistry

1 answer:

romanna [79]3 years ago

8 0

Answer:

When pH decreases acidity increases,

and if pH increases acidity decreases.

Explanation:

1 2 3 4 5 6 7(neutral) 8 9 10

acidic basic

So, when pH decreases acidity increases,

and if pH increases acidity decreases.

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What is the percent composition by mass of chlorine in LiCI

Lelechka [254]

Answer: Lithium (Li) 16.373%

Chlorine (Cl) 83.627%

Explanation:

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3 years ago

If the pka of hcho2 is 3.74 and the ph of an hcho2/nacho2 solution is 3.89, it means that

Andrei [34K]

Hi, the question has some missing part. The complete question should be " If the $pK_{a}$ of $HCHO_{2}$ is 3.74 and the pH of an $HCHO_{2}/NaCHO_{2}$ solution is 3.89, it means that -

(a) $[HCHO_{2}]=[NaCHO_{2}]$

(b) $[HCHO_{2}]> [NaCHO_{2}]$

(c) $[HCHO_{2}]< [NaCHO_{2}]$

Which of the options are true?"

Answer: Option (c) is true.

Explanation:

$HCHO_{2}/NaCHO_{2}$ is an weak acid-conjugate base buffer. Hence, in accordance with Henderson-Hasselbalch equation-

$pH=pK_{a}(HCHO_{2})+log(\frac{[CHO_{2}^{-}]}{[HCHO_{2}]})$

Here, pH = 3.89, $pK_{a}$ of $HCHO_{2}$ = 3.74

So, $3.89=3.74+log(\frac{[CHO_{2}^{-}]}{[HCHO_{2}]})$

or, $\frac{[CHO_{2}^{-}]}{[HCHO_{2}]}$ = 1.41 (>1)

So, $[CHO_{2}^{-}]> [HCHO_{2}]$

or, $[NaCHO_{2}]> [HCHO_{2}]$

Hence, option (c) is correct.

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4 years ago

When water vapor cools it condenses select the number that represents this process on the diagram

TiliK225 [7]

2 is the correct answer.

6 0

4 years ago

Read 2 more answers

A student observed an example in which all of the offspring of an organism have the same

Sphinxa [80]

The answer is C I believe because it makes sense

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3 years ago

Calculate the pH in titration of a weak acid: What is the pH in titration of formic acid (HCHO2, 0.200 M, 100.0 mL) after the ad

ki77a [65]

Answer:

pH = 12.61

Explanation:

First of all, we determine, the milimoles of base:

0.120 M = mmoles / 300 mL

mmoles = 300 mL . 0120 M = 36 mmoles

Now, we determine the milimoles of acid:

0.200 M = mmoles / 100 mL

mmoles = 100 mL . 0.200M = 20 mmoles

This is the neutralization:

HCOOH + OH⁻ ⇄ HCOO⁻ + H₂O

20 mmol 36 mmol 20 mmol

16 mmol

We have an excess of OH⁻, the ones from the NaOH and the ones that formed the salt NaHCOO, because this salt has this hydrolisis:

NaHCOO → Na⁺ + HCOO⁻

HCOO⁻ + H₂O ⇄ HCOOH + OH⁻ Kb → Kw / Ka = 5.55×10⁻¹¹

These contribution of OH⁻ to the solution is insignificant because the Kb is very small

So: [OH⁻] = 16 mmol / 400 mL → 0.04 M

- log [OH⁻] = pOH → 1.39

pH = 14 - pOH → 12.61

6 0

3 years ago