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Ugo [173]
2 years ago
13

Calculate [H+] for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral.

Chemistry
1 answer:
vesna_86 [32]2 years ago
3 0

Answer:

<em>Part A: </em>10.7.

<em>Part B: </em>2) the solution is basic.

<em>Part C: </em>5.9.

<em>Part D: </em>1) the solution is acidic.

<em>Part E: </em>8.1.

<em>Part F: </em>2) the solution is basic.

Explanation:

<em>Part A [OH−]= 4.5×10−4 M . Express your answer using two significant figures.  </em>

<em>∵ [H₃O⁺][OH⁻] = 10⁻¹⁴. </em>

<em>[OH⁻] = 4.5×10⁻⁴ M. </em>

<em>∴ [H₃O⁺] = 10⁻¹⁴/[OH⁻]  = 10⁻¹⁴/(4.5×10⁻⁴  M) = 2.22 × 10⁻¹¹ M.</em>

∵ pH = - log[H₃O⁺]

<em>∴ pH = - log(2.22 × 10⁻¹¹) = 10.65 ≅ 10.7.</em>

<em></em>

Part B 1) the solution is acidic 2) the solution is basic 3) the solution is neutral  

The solution is basic, because the pH is higher than 7.

<em>Part C [OH⁻]= 7.7×10⁻⁹ M . Express your answer using two significant figures.  </em>

<em>∵ [H₃O⁺][OH⁻] = 10⁻¹⁴. </em>

<em>[OH⁻] = 7.7×10⁻⁹ M. </em>

<em>∴ [H₃O⁺] = 10⁻¹⁴/[OH⁻]  = 10⁻¹⁴/(7.7×10⁻⁹ M) = 1.29 × 10⁻⁶ M.</em>

∵ pH = - log[H₃O⁺]

<em>∴ pH = - log(1.29 × 10⁻⁶) = 5.88 ≅ 5.9.</em>

<em>Part D 1) the solution is acidic 2) the solution is basic 3) the solution is neutral  </em>

The solution is acidic, because the pH is lower than 7.

<em>Part E A solution in which [OH−] is 140 times greater than [H+]. Express your answer using two significant figures.  </em>

<em>∵ [H₃O⁺][OH⁻] = 10⁻¹⁴. </em>

<em>∵  [OH⁻] = 140[H₃O⁺]</em>

<em>∴ [H₃O⁺](140[H₃O⁺]) = 10⁻¹⁴. </em>

<em>∴ 140 [H₃O⁺]² = 10⁻¹⁴. </em>

<em>[H₃O⁺]² = 10⁻¹⁴/ 140 = 7.14 x 10⁻¹⁷.</em>

<em>∴ [H₃O⁺] = √(7.14 x 10⁻¹⁷) = 8.45 x 10⁻⁹ M.</em>

<em></em>

∵ pH = - log[H₃O⁺]

<em>∴ pH = - log(8.45 x 10⁻⁹) = 8.07 ≅ 8.1.</em>

<em>Part F 1) the solution is acidic 2) the solution is basic 3) the solution is neutral</em>

<em></em>

The solution is basic, because the pH is higher than 7.

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