# Calculate [OH^-] each of the following solutions, and indicate whether the solution is acidic, basic, or neut [Solved] Since [H ] is 12 times greater than that, the solution is acidic

Since [H ] is 12 times greater than that, the solution is acidic

if [H ] = 8.510^-3, then
1.010^-14 = 8.510^-3[OH-]
[OH-] = 1.210^-12

right here equations are possibly important to be sure it truly is guy or woman-friendly/acidic pH= -log [H3O ] pOH= -log [OH-] pH pOH = 14 a) pH= -log[a million.0*10^-2]= 2 b) pOH= -log[a million.0*10^-2] = 2 yet it somewhat is pOH, no longer pH so 14-2= pH= 12 c) pOH= -log[a million.0*10^-8] = 8; pH= 6 d) pH= -log[a million.0*10^-6] = 6 something under 7 is seen acidic, something above 7 is guy or woman-friendly. So (a), (c), and (d) are acidic, and b could desire to be guy or woman-friendly

if [H ] = 7.510^-10, then
1.010^-14 = 7.510^-10[OH-]
[OH-] = 1.310^-5

if [H ] = 7.510^-10, then
1.010^-14 = 7.510^-10[OH-]
[OH-] = 1.310^-5

for #3, let [OH-] = x
1.010^-14 = (x)(12x)
1.010^-14 = 12x^2
x = 2.810^-8

Since [OH-] >> [H ], the solution is basic

Since [H ] is 12 times greater than that, the solution is acidic

This post is last updated on hrtanswers.com at Date : 1st of September – 2022

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