Calculate [h3o+] at 25 grades celsius when [oh-]=2.5 x 10 ^-9M
[H3O+]=Kw[OH−][H_3O^{+}]=\frac {K_w} {[OH^-] }[H3O+]=[OH−]Kw
[H3O+]=1.0×10−142.5×10−9[H_3O^+]=\frac {1.0×10 ^{-14}}{2.5×10 ^{-9}}[H3O+]=2.5×10−91.0×10−14
[H3O+]=4.0×10−6M[H_3O^+]=4.0×10 ^{-6}M[H3O+]=4.0×10−6M
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