1. **Problem statement:** We are given the formula for pH: $$\text{pH} = -\log [H^+]$$ where $[H^+]$ is the concentration of hydrogen ions in mol/L. We need to solve two parts: (a) Find $[H^+]$ when pH = 1.4. (b) Find the pH of a solution with 40 times the hydrogen ion concentration of a solution with pH = 9.5. --- 2. **Formula and rules:** - pH is defined as $$\text{pH} = -\log [H^+]$$ - To find $[H^+]$ from pH, rearrange: $$[H^+] = 10^{-\text{pH}}$$ - Logarithm properties: $$\log(a^b) = b \log a$$ and $$\log(10) = 1$$ --- 3. **Part (a): Find $[H^+]$ when pH = 1.4** $$[H^+] = 10^{-1.4}$$ Calculate: $$10^{-1.4} = 10^{-1 - 0.4} = 10^{-1} \times 10^{-0.4} = 0.1 \times 10^{-0.4}$$ Using approximate value $10^{-0.4} \approx 0.398$: $$[H^+] = 0.1 \times 0.398 = 0.0398$$ mol/L --- 4. **Part (b): Find pH of solution with 40 times more $[H^+]$ than solution with pH = 9.5** First, find $[H^+]$ for pH = 9.5: $$[H^+]_{9.5} = 10^{-9.5}$$ The new concentration is: $$[H^+]_{new} = 40 \times 10^{-9.5}$$ Find new pH: $$\text{pH}_{new} = -\log [H^+]_{new} = -\log(40 \times 10^{-9.5})$$ Using log properties: $$= - (\log 40 + \log 10^{-9.5}) = - (\log 40 - 9.5) = -\log 40 + 9.5$$ Calculate $\log 40$: $$\log 40 = \log (4 \times 10) = \log 4 + \log 10 = 0.602 + 1 = 1.602$$ So: $$\text{pH}_{new} = -1.602 + 9.5 = 7.898$$ --- **Final answers:** (a) $[H^+] = 0.0398$ mol/L (b) $\text{pH} = 7.898$