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Chemistry

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How many grams of KHP (204.22 g/mol) were present in container if 49.36 mL of 0.2264M NaOH were needed to reach the end point of a titration? I'm confused on the formula you would use for this.

  • Chemistry - ,

    Most problems in chemistry hinge on mols. A mole reacts with one/two/three/etc mols of some other material.

    KHP + NaOH ==> NaKP + H2O
    mols NaOH = M x L = ?
    molsl KHP = mols NaOH (from the coefficients in the balanced equation).
    Then mols KHP = grams KHP/molar mass KHP. You know molar mass and mols solve for grams KHP.

  • Chemistry - ,

    So, Mols NaOH = 0.2264 M NaOH x 0.04936 L NaOH = 0.011175
    Mols KHP = grams KHP / molar mass KHP or
    grams KHP = Mols KHP x molar mass KHP so
    0.011175 x 204.22 = 2.2822 grams of KHP
    Is that correct DBob222?

  • Chemistry - ,

    Yes, that is correct(almost); however, you have used too many significant figures. The 49.36 number limits the number of s.f. to four; therefore, you may have only four in the final answer. That means you should round your final answer to 2.282 g KHP.

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