$3.00 \cdot 10^{- 3}$ $3.00 * 10^-3$ mol ofHBr are dissolved in water to make 16 L of solution. What is the concentration of hydroxide ions, [ $O H^{-}$ $OH^-$ ] in this solution?

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$3.00 \cdot 10^{- 3}$ $3.00 * 10^-3$ mol ofHBr are dissolved in water to make 16 L of solution. What is the concentration of hydroxide ions, [ $O H^{-}$ $OH^-$ ] in this solution?

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Answer 1 · 2017-07-16T00:23:58

$[H O^{-}] = 5.33 \times 10^{- 11} \cdot m o l \cdot L^{- 1}$ $[HO^-]=5.33xx10^-11*mol*L^-1$

Explanation:

We know that $10^{- 14} = [H O^{-}] [H_{3} O^{+}]$ $10^-14=[HO^-][H_3O^+]$ under standard conditions at $298 \cdot K$ $298*K$ ......... And should we take ${log}_{10}$ $log_10$ of BOTH SIDES, we gets......

$underbrace(log_10(10^-14))_-14=log_10[H_3O^+] + log_10[HO^-]$

And on rearrangement, $14=underbrace(-log_10[H_3O^+])_(pH)underbrace(-log_10[HO^-])_(pOH)$

And so our defining relationship, $pH+pOH=14$

Now.............

$[H_3O^+]=(3.00xx10^-3*mol)/(16*L)=1.875xx10^-4*mol*L^-1$

$pH=-log_10(1.875xx10^-4)=-(-3.727)=3.727$

And thus $pOH=10.27$ , and......................

$[HO^-]=10^(-10.27)=5.33xx10^-11*mol*L^-1$

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$3.00 \cdot 10^{- 3}$ $3.00 * 10^-3$ mol ofHBr are dissolved in water to make 16 L of solution. What is the concentration of hydroxide ions, [ $O H^{-}$ $OH^-$ ] in this solution?

1 Answer

Explanation:

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3.00 * 10^-33.00⋅10−33.00 * 10^-3 mol ofHBr are dissolved in water to make 16 L of solution. What is the concentration of hydroxide ions, [OH^-OH−OH^-] in this solution?

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Explanation:

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$3.00 \cdot 10^{- 3}$ $3.00 * 10^-3$ mol ofHBr are dissolved in water to make 16 L of solution. What is the concentration of hydroxide ions, [ $O H^{-}$ $OH^-$ ] in this solution?