# Chapter 13 - Exercises - Page 575d: 93

It is necessary 0.163g of KOH.

#### Work Step by Step

1. Calculate the hydroxide ion concentration: pH + pOH = 14 11.56 + pOH = 14 pOH = 2.44 $[OH^-] = 10^{-pOH}$ $[OH^-] = 10^{- 2.44}$ $[OH^-] = 3.631 \times 10^{- 3}M$ - Since KOH is a strong base, $[KOH] = [OH^-] = 3.631 \times 10^{- 3}M$ 2. Calculate the number of moles: - 800ml = 0.8L $n(moles) = concentration(M) * volume(L)$ $n(moles) = 3.631\times 10^{- 3} * 0.8$ $n(moles) = 2.905\times 10^{- 3}$ 3. Find the mass value in grams: 39.1* 1 + 16* 1 + 1.01* 1 = 56.11g/mol $mass(g) = mm(g/mol) * n(moles)$ $mass(g) = 56.11 * 2.905\times 10^{- 3}$ $mass(g) = 0.163$

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