# What is the pOH of a 4.8xx10^(-10) M H^+ solution?

Mar 19, 2018

$p O H \equiv 4.68$

#### Explanation:

For aqueous solutions we know that $p H + p O H = 14$...

And so $p O H = 14 - p H$...

And since ...........

$p H = - {\log}_{10} \left[{H}^{+}\right] = - {\log}_{10} \left(4.8 \times {10}^{-} 10\right) = - \left(- 9.32\right) = + 9.32 \ldots .$

$p O H = 14 - 9.32 = 4.68$...
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}