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General Pharmaceutical Council Registration Assessment (GPhC Assessment) Pharmaceutical Calculations Flashcards
56 question-and-answer cards covering Pharmaceutical Calculations as it is examined in General Pharmaceutical Council Registration Assessment (GPhC Assessment). 24 of them are printed below, taken from across the deck — no signup, no paywall on the preview.
24 sample cards from the Pharmaceutical Calculations deck
Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.
Give the dilution relationship used to find the new concentration when a solution is diluted with solvent.
$$C_{1}V_{1}=C_{2}V_{2}$$ where $C_{1},V_{1}$ are the concentration and volume before dilution and $C_{2},V_{2}$ after. Solve for the unknown; the amount of drug is conserved.
How much water must be added to $50\text{ mL}$ of a $20\%\text{ w/v}$ solution to produce a $5\%\text{ w/v}$ solution?
$C_{1}V_{1}=C_{2}V_{2}\Rightarrow 20\times50=5\times V_{2}\Rightarrow V_{2}=200\text{ mL}$. Water to add $=200-50=150\text{ mL}$.
Define a 'dilution factor' and how it relates concentration before and after dilution.
The dilution factor is the ratio $\frac{V_{\text{final}}}{V_{\text{initial}}}$. The concentration is reduced by this factor: $C_{\text{final}}=\frac{C_{\text{initial}}}{\text{dilution factor}}$. E.g. diluting $1\text{ mL}$ to $10\text{ mL}$ is a factor of 10 and gives one-tenth the concentration.
What is the alligation method used for in pharmaceutical calculations?
Alligation (alligation alternate) is a technique to determine the proportions in which two preparations of different strengths must be mixed to obtain a product of a desired intermediate strength.
Using alligation, in what ratio should a $10\%$ and a $2\%$ cream be mixed to make a $4\%$ cream?
Differences across the desired strength: $10\%$ side gives $4-2=2$ parts; $2\%$ side gives $10-4=6$ parts. So mix $10\%:2\%=2:6=1:3$ (1 part of the $10\%$ to 3 parts of the $2\%$).
How can you verify an alligation result by checking the amount of active ingredient?
Compute total active from each component and divide by total quantity. For $1$ part $10\%$ + $3$ parts $2\%$ (say $100\text{ g}$ each gives $1\times10 + 3\times2 = 16\text{ g}$ active in $400\text{ g}$): $\frac{16}{400}\times100=4\%$, confirming the target strength.
What is a master formula, and what does scaling a master formula involve?
A master formula lists the quantities of each ingredient to make a stated total quantity of a preparation. Scaling multiplies every ingredient by the same factor $\frac{\text{required quantity}}{\text{formula quantity}}$ so the proportions are preserved.
A formula makes $10$ suppositories but you need $30$. By what factor do you multiply each ingredient, and why prepare a small excess?
Scaling factor $=\frac{30}{10}=3$, so multiply every ingredient by 3. An excess (e.g. calculate for a few extra units) is prepared to allow for unavoidable losses during mixing, moulding and transfer so the required number can actually be produced.
What is a displacement value (DV) in suppository/pessary preparation?
The displacement value is the number of parts by weight of a drug that displaces 1 part by weight of the base. It accounts for the fact that a given mass of drug occupies the space of a different mass of base in the mould.
State the formula for the mass of base required when making medicated suppositories using a displacement value.
$$\text{Base required}=\big(\text{No. of supps}\times\text{mould calibration}\big)-\frac{\text{total drug mass}}{\text{DV}}$$ where mould calibration is the mass of pure base that fills one mould.
For a drug with DV $=2$, how much base is displaced by $400\text{ mg}$ of drug?
Displaced base $=\frac{\text{drug mass}}{\text{DV}}=\frac{400}{2}=200\text{ mg}$ of base. So $400\text{ mg}$ of this drug takes the place of $200\text{ mg}$ of base in the mould.
What is the basic formula linking infusion rate, volume and time?
$$\text{Rate (mL/h)}=\frac{\text{volume (mL)}}{\text{time (h)}}$$ For a given volume to be infused over a set time, divide volume by time to get the pump rate.
How do you convert an infusion rate in mL/h to drops per minute using a giving set's drop factor?
$$\text{Drops/min}=\frac{\text{volume (mL)}\times\text{drop factor (drops/mL)}}{\text{time (min)}}$$ Common drop factors are $20\text{ drops/mL}$ (standard solution set) and $60\text{ drops/mL}$ (microdrop/paediatric set).
A patient needs a drug at $4\ \text{mg/min}$ from a bag containing $2\text{ g}$ in $500\text{ mL}$. What infusion rate in mL/h is required?
Concentration $=\frac{2000\text{ mg}}{500\text{ mL}}=4\text{ mg/mL}$. Required flow $=\frac{4\text{ mg/min}}{4\text{ mg/mL}}=1\text{ mL/min}=60\text{ mL/h}$.
How do you calculate a weight-based infusion rate in mL/h (e.g. $\mu$g/kg/min) for a given concentration?
$$\text{Rate (mL/h)}=\frac{\text{dose }(\mu\text{g/kg/min})\times\text{weight (kg)}\times60}{\text{concentration }(\mu\text{g/mL})}$$ Multiply dose by weight to get $\mu$g/min, by 60 to get $\mu$g/h, then divide by the bag concentration.
Why must the displacement (reconstitution) volume of a dry powder injection be accounted for, and how does it affect final concentration?
The powder itself occupies volume, so the final volume exceeds the diluent added. Final concentration $=\frac{\text{drug amount}}{\text{final volume}}$, where final volume = diluent volume + powder displacement. Ignoring displacement underestimates the volume and overestimates concentration, giving the wrong dose volume.
A vial contains $1\text{ g}$ powder with a displacement volume of $0.8\text{ mL}$. If $9.2\text{ mL}$ of water is added, what is the final concentration?
Final volume $=9.2+0.8=10\text{ mL}$. Concentration $=\frac{1000\text{ mg}}{10\text{ mL}}=100\text{ mg/mL}$.
In pharmacokinetics, define volume of distribution and give its formula.
Volume of distribution ($V_{d}$) is the apparent volume into which a drug distributes to give the observed plasma concentration. $$V_{d}=\frac{\text{amount of drug in body}}{C_{p}}$$ where $C_{p}$ is the plasma concentration; it relates dose to concentration and can be used to estimate a loading dose.
State the relationship between elimination rate constant, half-life and clearance.
$$t_{1/2}=\frac{0.693}{k}\quad\text{and}\quad CL=k\times V_{d}$$ where $k$ is the first-order elimination rate constant, $t_{1/2}$ the half-life, $CL$ clearance and $V_{d}$ volume of distribution ($0.693=\ln 2$).
How are loading dose and maintenance dose calculated in pharmacokinetics?
$$\text{Loading dose}=\frac{C_{\text{target}}\times V_{d}}{F}\qquad \text{Maintenance dose rate}=\frac{C_{\text{target}}\times CL}{F}$$ where $C_{\text{target}}$ is the desired steady-state concentration, $F$ bioavailability, $V_{d}$ volume of distribution and $CL$ clearance.
What is a milliequivalent (mEq), and how is it calculated from millimoles and valency?
A milliequivalent measures the chemical combining capacity of an ion. $$\text{mEq}=\text{mmol}\times\text{valency}$$ For monovalent ions (e.g. $\ce{Na+}$, $\ce{K+}$) $1\text{ mmol}=1\text{ mEq}$; for divalent ions (e.g. $\ce{Ca^2+}$) $1\text{ mmol}=2\text{ mEq}$.
Calcium chloride dihydrate has $M=147\ \text{g mol}^{-1}$. How many millimoles of $\ce{Ca^2+}$ and how many mEq are in $1.47\text{ g}$?
$\text{mmol}=\frac{1470\text{ mg}}{147}=10\text{ mmol of }\ce{Ca^2+}$. As $\ce{Ca^2+}$ is divalent, $\text{mEq}=10\times2=20\text{ mEq}$.
How do you convert a sodium content in mmol to a mass of sodium chloride, and why does this matter clinically?
Each mmol of $\ce{Na+}$ comes from $1\text{ mmol }\ce{NaCl}=58.5\text{ mg}$. So mass $\ce{NaCl}=\text{mmol Na}\times58.5\text{ mg}$. This matters for monitoring sodium load in patients on restricted intake (e.g. heart failure, renal disease) receiving IV fluids or effervescent preparations.
What is osmolarity, and how does it differ for a salt that dissociates into multiple ions?
Osmolarity is the number of osmotically active particles (osmoles) per litre of solution. A salt contributes osmoles equal to moles $\times$ number of ions it dissociates into; e.g. $1\text{ mmol }\ce{NaCl}$ gives $2\text{ mosmol}$ ($\ce{Na+}+\ce{Cl-}$), whereas $1\text{ mmol}$ of glucose gives $1\text{ mosmol}$.
What this deck covers
The Pharmaceutical Calculations deck follows the General Pharmaceutical Council Registration Assessment (GPhC Assessment) Pharmaceutical Calculations syllabus — 4 chapters and 16 topics — so questions land on material that is genuinely examinable rather than trivia around it. That works out to roughly 14.0 cards per chapter.
Answers are written to be recallable, not just readable — averaging about 223 characters, which is long enough to carry the reasoning and short enough to say out loud.
A deck like this earns its keep on the second and third pass. Read the syllabus first so you know the shape of the subject, then use the cards to find the specific facts that have not stuck.
Pharmaceutical Calculations flashcards FAQ
How many Pharmaceutical Calculations flashcards are in this General Pharmaceutical Council Registration Assessment (GPhC Assessment) deck?
56 cards. This page previews 24 of them, sampled evenly across the deck so you can judge the difficulty before installing anything.
Are these General Pharmaceutical Council Registration Assessment (GPhC Assessment) flashcards free?
Yes. The preview here is free to read with no signup, and the full 56-card deck is free inside the Examius app.
What do the Pharmaceutical Calculations cards cover?
They follow the General Pharmaceutical Council Registration Assessment (GPhC Assessment) Pharmaceutical Calculations syllabus — 4 chapters and 16 topics — so the questions track what is actually examinable.
How should I use these flashcards?
Read the syllabus first so you know the shape of the subject, then drill the deck. Examius schedules each card with spaced repetition, so cards you keep missing come back sooner and ones you know drift further apart.