# Chapter 17 - Additional Aspects of Acid-Base Equilibria - Exercises - Neutralization Reactions - Page 777: 35

$C_2 (H_3PO_4) = 0.1508M$

#### Work Step by Step

1000ml = 1L 31.15ml = 0.03115 L 25ml = 0.025 L 1. Find the concentration necessary for an equal number of moles: Notice: Since $H_3PO_4$ is capable of giving a proton 2 times, the number of moles of $KOH$ has to be 2 times the number for the acid: $(Base)(C_1 * V_1) \div 2= (Acid)V_2 * C_2$ $(0.242* 0.03115) \div 2= 0.025 * C_2$ $0.003769 = 0.025 * C_2$ $C_2 = 0.1508M$

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