To calculate the concentration of \([H_3O^+]\) in an aqueous solution where the concentration of hydroxide ions, \([OH^-]\), is given, we can use the relationship defined by the ion product of water:
\[ K_w = [H_3O^+][OH^-] \]
At 25°C, the ion product of water (\(K_w\)) is:
\[ K_w = 1.0 \times 10^{-14} \]
Given that the concentration of hydroxide ions \([OH^-] = 1.8 \times 10^{-10} , M\), we can rearrange the equation to solve for \([H_3O^+]\):
\[ [H_3O^+] = \frac{K_w}{[OH^-]} \]
Substituting in the values:
\[ [H_3O^+] = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-10}} \]
Now we perform the calculation:
\[ [H_3O^+] = \frac{1.0}{1.8} \times 10^{-14 + 10} = \frac{1.0}{1.8} \times 10^{-4} \]
Calculating \(\frac{1.0}{1.8}\) gives approximately 0.5555, so we multiply:
\[ [H_3O^+] \approx 0.5555 \times 10^{-4} = 5.555 \times 10^{-5} , M \]
Rounding to two significant figures, we get:
\[ [H_3O^+] \approx 5.6 \times 10^{-5} , M \]
Therefore, the concentration of \([H_3O^+]\) in the solution is:
\[ \boxed{5.6 \times 10^{-5} , M} \]
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