How many mL of a .10 M $N a O H$ $NaOH$ solution are needed to neutralize 15 mL of .20 M $H_{3} P O_{4}$ $H_3PO_4$ solution?

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Answer 1 · 2017-06-19T21:48:58

30 mL

Use the formula $M_{A} V_{A} = M_{B} V_{B}$ $M_AV_A = M_BV_B$ .

$M_{A} =$ $M_A =$ Molarity of acid, $V_{A} =$ $V_A =$ Volume of acid
$M_{B} =$ $M_B =$ Molarity of base, $V_{B} =$ $V_B =$ Volume of base

1. Plug known values into the formula.

$(.20) (15) = .10 (x)$ $(.20)(15) = .10(x)$

2. Solve for x.

$3 = .10 x$ $3 = .10x$
$x = 30$ $x = 30$

30 mL is your answer.

How many mL of a .10 M NaOHNaOHNaOH solution are needed to neutralize 15 mL of .20 M H_3PO_4H3PO4H_3PO_4 solution?