What is the $p H$ $pH$ of a mass of $4.66 \cdot m g$ $4.66mg$ of barium hydroxide dissolved in a $1 \cdot L$ $1L$ volume of water?

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What is the $p H$ $pH$ of a mass of $4.66 \cdot m g$ $4.66mg$ of barium hydroxide dissolved in a $1 \cdot L$ $1L$ volume of water?

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Answer 1 · 2016-10-26T19:23:10

$p H = 9.74$ $pH=9.74$

Explanation:

We need to find (i), $[H O^{-}]$ $[HO^-]$ $=$ $=$ $\frac{2 \times 4.66 \times 10^{- 3} g}{171.34 \cdot g \cdot m o l^{- 1}} \times \frac{1}{L}$ $(2xx4.66xx10^-3g)/(171.34*g*mol^-1)xx1/L$ $=$ $=$ $5.44 \times 10^{- 5} m o l \cdot L^{- 1}$ $5.44xx10^-5mol*L^-1$ .

And (ii), $p O H$ $pOH$ $=$ $=$ $- {log}_{10} [H O^{-}]$ $-log_10[HO^-]$ $=$ $=$ $- {log}_{10} (5.44 \times 10^{- 5})$ $-log_10(5.44xx10^-5)$ $=$ $=$ $4.26 .$ $4.26.$

And since (iii) $p H + p O H = 14, p H = 14 - 4.26 = 9.74$ $pH+pOH=14, pH=14-4.26=9.74$ .

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What is the $p H$ $pH$ of a mass of $4.66 \cdot m g$ $4.66mg$ of barium hydroxide dissolved in a $1 \cdot L$ $1L$ volume of water?

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What is the pHpH of a mass of 4.66⋅mg4.66*mg of barium hydroxide dissolved in a 1⋅L1*L volume of water?

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What is the $p H$ $pH$ of a mass of $4.66 \cdot m g$ $4.66mg$ of barium hydroxide dissolved in a $1 \cdot L$ $1L$ volume of water?