Isotope dilution:a mass spectrometric technique in which a known amount of an unusual isotope (called the spike) is added to an un

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Isotope dilution:a mass spectrometric technique in which a known amount of an unusual isotope (called the spike) is added to an unknown as an internal standard for quantitative analysis.

Isotope dilution is a mass spectrometric technique in which a known amount of an unusual isotope (called the spike) is added to an unknown as an internal standard for quantitative analysis. The ratio of the isotopes is measured and from this ratio the quantity of the element in the original unknown can be calculated. For example, natural vanadium atom fractions 51V = 0.9975 and 50V = 0.0025. The atomic fraction is defined as: atom fraction of 51V = atoms 51V/(atoms 51V + atoms 50 V) A spike enriched in 50 V has atom fractions 51V = 0.6391 and 50V = 0.3609. a) Let A be 51V and B be 50V. Let Ax be the atom fraction of A in an unknown. Let Bx be the atom fraction of B in the unknown. Let As and Bs be the corresponding atom fractions in a spike. Let Cx be the total concentration of all isotopes of vanadium (Pmol/g) in the unknown, and let Cs be the concentration in the spike. After mixing mx grams of unknown with ms grams of the spike, show that the ratio of isotopes in the mixture (denoted R) is given by � = ��� � ��� � = �௫�௫�௫ + �௦�௦�௦ �௫�௫�௫ + �௦�௦�௦ b) Solve the above equation for Cx to show that �௫ = ൬ �௦�௦ �௫ ൰ ൬ �௦ − ��௦ ��௫ − �௫ ൰ c) A 0.40167 g sample of crude oil containing an unknown concentration of natural vanadium was mixed with a 0.41946 g spike containing 2.2435 Pmol/g enriched with 50V (atom fractions: 51V = 0.6391 and 50V = 0.3609). The measured isotope ratio by mass spectrometry was R = 51V /50V = 10.545. Determine the concentration of vanadium (Pmol/g) in the crude oil. 6. (15 pts.) Consider the electrochemical cell described below. ��(�)|����(�)|���(��, ���������)||���� ��������|��(�) The cell solution was made by mixing 25.0 mL of 4.00 mM ��� 25.0 mL of 4.00 mM ���(��)ଶ 25.0 mL of 0.400 M acid, ��, with pKa = 9.50 25.0 mL of ��� solution The measured voltage was -0.440 V. Calculate the molarity of the KOH solution. Assume that essentially all of the copper(I) is in the form ��(��)ଶ ି. For the right side (the cathode), the half cell reaction is: ��(��)ଶ ି + �ି ⇌ ��(�) + 2��ି �° = −0.429 �  8. (15 pts.) An ion selective electrode used to measure ��ଶା is also sensitive to ��ା and obeys the equation � = �������� + 57.1 �� 2 log{[��ଶା] + 0.002[��ା]ଶ} When the electrode was immersed in 100.0 mL of unknown containing ��ଶା in 0.200 � ����, the reading was 194.6 ��. When 1.00 �� of 1.07 × 10ିଶ � ��ଶା (in 0.200 � ���� ) was added to the unknown, the reading increased to 200.7 ��. Find the concentration of ��ଶା in the original unknown.