If the H+ concentration of pure water is increased to ten times the original concentration, what will be the resulting OH- concentration?

If the H+ concentration of pure water is increased to ten times the original concentration, what will be the resulting OH- concentration?

• A. 10-6 M
• B. 10-7 M
• C. 10-8 M
• D. 10-9 M

Answer: (C)

In pure water, the concentration of hydrogen ions (H+) and hydroxide ions (OH-) is equal, typically around 1 x 10^-7 M at 25°C. This results in a neutral pH of 7. When the H+ concentration is increased to ten times the original amount, it reaches 1 x 10^-6 M. In water, the product of H+ and OH- concentrations is constant at 1 x 10^-14 M at 25°C, represented by the formula: \[ [H^+] \times [OH^-] = 1 \times 10^{-14} \] Given the new concentration of H+ (1 x 10^-6 M), you can rearrange this equation to solve for the concentration of OH-: \[ [OH^-] = \frac{1 \times 10^{-14}}{[H^+]} \] \[ [OH^-] = \frac{1 \times 10^{-14}}{1 \times 10^{-6}} = 1 \times 10^{-8} M \] Thus, increasing the H+ concentration to ten times results in an OH- concentration of 1 x 10^-8 M

In pure water, the concentration of hydrogen ions (H+) and hydroxide ions (OH-) is equal, typically around 1 x 10^-7 M at 25°C. This results in a neutral pH of 7. When the H+ concentration is increased to ten times the original amount, it reaches 1 x 10^-6 M.

In water, the product of H+ and OH- concentrations is constant at 1 x 10^-14 M at 25°C, represented by the formula:

[ [H^+] \times [OH^-] = 1 \times 10^{-14} ]

Given the new concentration of H+ (1 x 10^-6 M), you can rearrange this equation to solve for the concentration of OH-:

[ [OH^-] = \frac{1 \times 10^{-14}}{[H^+]} ]

[ [OH^-] = \frac{1 \times 10^{-14}}{1 \times 10^{-6}} = 1 \times 10^{-8} M ]

Thus, increasing the H+ concentration to ten times results in an OH- concentration of 1 x 10^-8 M