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1. Write the balanced chemical equation for the reaction
NaOH + HCI --> NaCI + H20

2. Extract the relevant information from the qustion:
NaOH v= 30mL, M=0.10 HCI v= 25.0 mL, M=?

3. Convert to Liters
NaOH v= 0.03 L, M= 0.10M HCI v=0.025L, M=?

4. Calculate moles NaOh:
n(NaOH) = M x V = 0.10 moles/L x 0.03L = 3x10^-3 moles

5. From the balanced chemical equation find the mole ratio
NaOH:HCI = 1:1

6. Find the number of moles of HCI that were titrated
NaOH: HCI is 1:1;
so n(NaOH) = n(HCI) = 0.003 moles at neutralization

7. Calculate the concentration of HCI:
M=n divided by V n=0.003 mol, V=0.025L

  • Actually inorganic chemistry -

    I will take a look but am no chemist.

  • Math -

    NaOH + HCI --> NaCI + H20
    Na balanced
    O balanced
    H balanced, 2 each side
    Cl balanced
    So it is balanced as is.

  • Math -

    NaOH v= 30mL, M=0.10 HCI v= 25.0 mL, M=?

    You need exactly the same number of moles (or molecules) of NaOH as of HCl
    .1 * 30 = M * 25
    M of HCl = .1 (30/25) = .12

  • Math -

    3. Convert to Liters
    NaOH v= 0.03 L, M= 0.10M HCI v=0.025L, M=?

    The same answer as #2

  • Math -

    I agree with 4 through 7

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