Calculating pH from hydrogen ion concentration and significant figure note

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Calculating pH from hydrogen ion concentration and significant figure note: To calculate the pH from the hydrogen ion concentration, take the negative log of the molarity of hydrogen ion. This may be expressed as [H+] or [H3O+]. If the acid is strong, the hydrogen ion concentration can be assumed to be the same as the molarity of the acid. The pH value should be expressed so that the number of digits to the right of the decimal point is equal to the number of significant figures in the hydrogen ion concentration. Use the log function for these calculations. Do not use the In function. This is for natural logs! Sample calculation: Suppose we want the pH of a solution that has an [H+] = 3.5×10^-7. On a typical scientific calculator, the following sequence would be entered: (-) log 3.5 EE (-) 7. The result displayed is 6.455931956. Since the original value had two significant figures, this would be rounded to 6.46, which is 2 digits to the right of the decimal. Acid Concentration PH Pure water with [H3O+] = 1.0×10^-7 0.00066 M HCl [H2O] = 3.86×10^5 1.88×10^-2 M HNO2 0.00555 M HBr

The Correct Answer and Explanation is:

To calculate the pH from the hydrogen ion concentration [H+][H^+][H+], we use the following formula: pH=−log⁡[H+]\text{pH} = -\log[H^+]pH=−log[H+]

This calculation involves using the base-10 logarithm (log) function, and the result should reflect the significant figures in the given hydrogen ion concentration. Let’s break down each part of your example:

1. Pure Water with [H3O+]=1.0×10−7[H_3O^+] = 1.0 \times 10^{-7}[H3​O+]=1.0×10−7 M

  • For pure water, the concentration of [H3O+][H_3O^+][H3​O+] is 1.0×10−71.0 \times 10^{-7}1.0×10−7 M, so: pH=−log⁡(1.0×10−7)=7\text{pH} = -\log(1.0 \times 10^{-7}) = 7pH=−log(1.0×10−7)=7 Since the concentration had 2 significant figures, the pH value, 777, reflects this as a whole number, and no decimal places are required.

2. 0.00066 M HCl

  • Hydrochloric acid (HCl) is a strong acid, meaning it dissociates completely in solution. Therefore, [H+]=0.00066[H^+] = 0.00066[H+]=0.00066 M. pH=−log⁡(0.00066)≈3.779\text{pH} = -\log(0.00066) \approx 3.779pH=−log(0.00066)≈3.779 The concentration 0.000660.000660.00066 has 2 significant figures, so the pH is rounded to 2 decimal places: pH≈3.78\text{pH} \approx 3.78pH≈3.78

3. [H3O+]=3.86×105[H_3O^+] = 3.86 \times 10^5[H3​O+]=3.86×105 for 1.88×10^-2 M HNO2

  • Nitrous acid (HNO2) is a weak acid, so its dissociation is not complete. We would need more information (like the acid dissociation constant) to calculate the exact [H+][H^+][H+], but if we assume complete dissociation (just for this example): [H+]=1.88×10−2[H^+] = 1.88 \times 10^{-2}[H+]=1.88×10−2 pH=−log⁡(1.88×10−2)≈1.726\text{pH} = -\log(1.88 \times 10^{-2}) \approx 1.726pH=−log(1.88×10−2)≈1.726 Since the concentration has 3 significant figures, we round the pH value to 3 decimal places: pH≈1.73\text{pH} \approx 1.73pH≈1.73

4. 0.00555 M HBr

  • Hydrobromic acid (HBr) is another strong acid, so it dissociates completely. Hence, [H+]=0.00555[H^+] = 0.00555[H+]=0.00555 M. pH=−log⁡(0.00555)≈2.255\text{pH} = -\log(0.00555) \approx 2.255pH=−log(0.00555)≈2.255 The concentration 0.005550.005550.00555 has 3 significant figures, so the pH is rounded to 3 decimal places: pH≈2.26\text{pH} \approx 2.26pH≈2.26

Summary:

  • Pure Water: pH = 7 (since [H3O+][H_3O^+][H3​O+] = 1.0×10−71.0 \times 10^{-7}1.0×10−7)
  • 0.00066 M HCl: pH = 3.78 (rounded to 2 decimal places)
  • 1.88×10^-2 M HNO2: pH = 1.73 (rounded to 2 decimal places)
  • 0.00555 M HBr: pH = 2.26 (rounded to 2 decimal places)

Each pH value corresponds to the given concentration, and the number of decimal places matches the number of significant figures in the concentration used for the calculation.

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