Answer
$C_2 (H_3PO_4) = 0.1508M$
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$
$C_2 = 0.1508M$
