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3 years ago

11

He table below lists some acids and bases. Strong acids Weak acids Weak bases Strong bases HBr CH3COOH NH3 NaOH HNO3 HCOOH NH4OH

KOH H2SO4 Ca(OH)2 Which salt has a pH < 7?
A) KBr
B) HCOOK
C) CaSO4
D) NH4NO3

2 answers:

ZanzabumX [31]3 years ago

3 0

The answer is D)NH4NO3

goldenfox [79]3 years ago

3 0

<h2>Answer : Option D) $NH_{4}NO_{3}$ </h2><h3>Explanation :</h3>

Amongst the given list of acids and bases, following are the strong acids and weak acids

Strong Acid: HBr, $HNO_{3}$ and $H_{2}SO_{4}$

Weak Acid: $CH_{3}COOH$ and HCOOH

And strong bases with weak bases are given below;

Strong Base: NaOH, KOH and $Ca(OH)_{2}$

Weak Base: $NH_{4}OH$ and $NH_{3}$

Now, Acid + Base → Salt + Water

so, the salt of strong acid and weak base which will have pH < 7

will be $NH_{4}NO_{3}$

because of the following acid base reaction (neutralization reaction)

$HNO_{3}$ + $NH_{4}OH$ ↔ $NH_{4}NO_{3}$ + $H_{2}O$

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3 years ago

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4 years ago

If you mix 50mL of 0.1 M TRIS acid with 60 mL of0.2 M<br> TRIS base, what will be the resulting pH?

Katyanochek1 [597]

<u>Answer:</u> The pH of resulting solution is 8.7

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

<u>For TRIS acid:</u>

Molarity of TRIS acid solution = 0.1 M

Volume of solution = 50 mL

Putting values in above equation, we get:

$0.1M=\frac{\text{Moles of TRIS acid}\times 1000}{50mL}\\\\\text{Moles of TRIS acid}=0.005mol$

<u>For TRIS base:</u>

Molarity of TRIS base solution = 0.2 M

Volume of solution = 60 mL

Putting values in above equation, we get:

$0.2M=\frac{\text{Moles of TRIS base}\times 1000}{60mL}\\\\\text{Moles of TRIS base}=0.012mol$

Volume of solution = 50 + 60 = 110 mL = 0.11 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[salt]}{[acid]})$

$pH=pK_a+\log(\frac{[\text{TRIS base}]}{[\text{TRIS acid}]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of TRIS acid = 8.3

$[\text{TRIS acid}]=\frac{0.005}{0.11}$

$[\text{TRIS base}]=\frac{0.012}{0.11}$

pH = ?

Putting values in above equation, we get:

$pH=8.3+\log(\frac{0.012/0.11}{0.005/0.11})\\\\pH=8.7$

Hence, the pH of resulting solution is 8.7

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4 years ago