Jasmine asked in Science & MathematicsChemistry · 10 years ago

Can you please help me with my Chemistry?

How many grams of Mg(OH)2 will be needed to neutralize 25 mL of stomach acid if stomach acid is 0.10 M HCl

How many mL of a 0.10 M NaOH solution are needed to neutralize 15 mL of 0.20 M H3PO4 solution?

Can you please help me explain how you did it and what the answer is because I don't understand.

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  • 10 years ago
    Best answer

    I can help with the second question:

    How many mL of a 0.10 M NaOH solution are needed to neutralize 15 mL of 0.20 M H3PO4 solution?

    use the formula MaVa=nb/naMbVb

    Ma- Molarity of the acid

    Va- Volume of the acid

    nb- number of moles of the base

    na- number of moles of the acid

    Mb-Molarity of the base

    Ma- Molarity of the acid

    In your question, Mb, Ma and Va are given (NaOH is the base, H3PO4 is the acid). First, before plugging the values in the formula to solve for Volume of the acid, you want to write out the reaction:

    NaOH + H3PO4 ---> H2O + NaH2PO4

    Then you want to balance the equation. Here, the equation is balanced, so the number of moles for the base (nb) is 1 and the number of moles of the acid (na) is 1.

    Now that you have every value except for Va, plug in the values in the formula and solve for Va.

    MaVa= nb/naMbVb

    (.1M)Va=(1/1)(.2M)(15mL)

    (.1M)Va=(.2M)(15mL)

    (.1M)Va=3MmL

    divide both sides by .1M

    Va=30mL

    Since you need all the values for the formula except for what you're solving for, this formula can't work for the first question. I'm not sure how to solve that question, but at least you're started off on the second question. most of the time, chemistry is just plugging in values in a formula that has all the values except the value you're looking for.

  • 10 years ago

    First question:

    The reaction between magnesium hydroxide and the "stomach acid" can be represented by the following equation:

    Mg(OH)2(s) + 2 HCl(aq) → MgCl2(aq) + 2 H2O(l)

    First, we need to find the number of moles of "stomach acid" present in 25 mL of 0.10 M HCl. We can use the following equation:

    moles HCl = (molarity HCl)(Liters HCl)

    moles HCl = (0.10 M)(0.025 L)

    moles HCl = 0.0025 mol

    Next, we can use dimensional analysis and equation coefficients to convert 0.0025 mole of hydrochloric acid to moles of magnesium hydroxide, to grams of magnesium hydroxide. We shall need the following equalities to set up conversion factors:

    2 mol HCl = 1 mol Mg(OH)2

    1 mol Mg(OH)2 = 58.319 g Mg(OH)2

    [(0.0025 mol HCl)/1][(1 mol Mg(OH)2)/(2 mol HCl)][(58.319 g Mg(OH)2)/(1 mol Mg(OH)2)] = 0.0728987 g Mg(OH)2 or 0.73 g Mg(OH)2 rounded to two significant figures

    Answer: About 0.73 grams of Mg(OH)2 will be needed to neutralize 25 mL of stomach acid if stomach acid is 0.10 M HCl.

    Second question:

    The reaction between sodium hydroxide solution and the phosphoric acid solution can be represented by the following equation:

    3 NaOH(aq) + H3PO4(aq) → Na3PO4(aq) + 3 H2O(l)

    Since the molar ratio between sodium hydroxide and phosphoric acid is not 1:1, but 3:1, we need to use the following equation:

    (normality of NaOH)(volume of NaOH) = (normality of H3PO4)(volume H3PO4)

    normality of NaOH = 0.10 N

    volume of NaOH = ?

    normality of H3PO4 = 3 × 0.20 = 0.60 N

    volume of H3PO4 = 15 mL

    volume of NaOH = [(normality of H3PO4)(volume H3PO4)] / (normality of NaOH)

    volume of NaOH = [(0.60 N)(15 mL)] / (0.10 N)

    volume of NaOH = 90. mL

    Answer: 90. mL of a 0.10 M NaOH solution are needed to neutralize 15 mL of 0.20 M H3PO4 solution.

    Source(s): Periodic Table of the Elements personal knowledge
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